aonecoding
BAN USER 2of 2 votes
AnswersGiven a binary tree, find the closest LEAF node to the target.
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Amazon Software Engineer  6of 6 votes
AnswersGive an array A of n integers where 1 <= a[i] <= K.
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Find out the length of the shortest sequence that can be constructed out of numbers 1, 2, .. k that is NOT a subsequence of A.
eg. A = [4, 2, 1, 2, 3, 3, 2, 4, 1], K = 4
All single digits appears. Each of the 16 double digit sequences, (1,1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2) ... appears. Because (1, 1, 2) doesn't appear, return 3. Report Duplicate  Flag  PURGE
Google Solutions Engineer  1of 1 vote
AnswerGiven an undirected graph represented as a list of edges, find out the number of connected component.
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Twitter Software Engineer  2of 2 votes
AnswersGive m balls and n bins. Find out how many ways to assign balls to bins. Notice the buckets has no order. Like (1,2,3) and (3,2,1) are considered the same.
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eg, m = 3, n = 2, return 2. (1, 2) and (3, 0) Report Duplicate  Flag  PURGE
Amazon Software Developer  2of 2 votes
AnswersGive an positive integer n, find out the smallest integer m, such that all digits in m multiply equals to n. For example, n = 36, return 49. n = 72, return 89. You can assume there is no overflow.
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Microsoft Software Engineer  2of 2 votes
AnswersYahoo Sunnyvale onsite
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A string s3 consists of multiple repetitions of s1.
Given s1 and another string s2, find if s2 is a substring of s3.
s3 = s1 + s1 + … + s1 = n * s1, where n is a positive integer 0.
For example
s1 = “aabc”, s2 = “caa” => true
s1 = “aabc”, s2 = “cab” => false
s1 = “aabc”, s2 = “caabcaa” => true Report Duplicate  Flag  PURGE
Yahoo Software Engineer Algorithm  5of 5 votes
AnswersCongrats on aonecode member A.P. for signing the offer with FB! Thanks for sharing the experience with us.
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phone:
postorder tree traversal recursive > iterative
add two binary number
onsite:
1 ring buffer
2 merge intervals
3 Leetcode alien dictionary
4.sort list of words Report Duplicate  Flag  PURGE
Facebook Software Engineer Algorithm  2of 2 votes
AnswerAirbnb: Design webbrowser back button
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Your web browser supports will support three actions: back, forward and open. The init webpage is “about:blank”.
Given a sequence of commands. Return the result page. Report Duplicate  Flag  PURGE
Airbnb Software Engineer Algorithm  2of 2 votes
AnswersApril Google Interview 3/4
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Maze
N,M array
Level 1 0,0 to N1,M1 => Path exsits?
Level 2 if path exists then print path
Level 3 if path exists then print min cost path
Level 4 O(nm log mn) using Priority Queue. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  6of 6 votes
AnswersFB Onsite March
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Q: Find number of Islands.
XXXOO
OOOXX
XXOOX
Return 3 islands.
1 1 1OO
OOO2 2
3 3OO 2
Followup: If the board is too big to fit in memory, how to get the number? Report Duplicate  Flag  PURGE
Facebook Software Engineer Algorithm  2of 2 votes
AnswersDropbox
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Grid Illumination: Given an NxN grid with an array of lamp coordinates.
Each lamp provides illumination to every square on their x axis,
every square on their y axis, and every square that lies in their diagonal
(think of a Queen in chess).
Given an array of query coordinates,
determine whether that point is illuminated or not. The catch is when checking a query all lamps adjacent to, or on,… Report Duplicate  Flag  PURGE
Dropbox Software Engineer Algorithm  2of 2 votes
AnswersMarch 2018 Phone Interview FB
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Calculate a moving average that considers the last N values.
Circular Queue (Interviewer didn't agree with the linked list queue that I suggested at first. Said the pointers took space) Report Duplicate  Flag  PURGE
Facebook Software Engineer Algorithm  4of 4 votes
AnswersFeb Onsite Google
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DP Problem. Given the length and width of a matrix, get the number of paths from bottomleft to bottom right.
You may only walk into those 3 directions ➡ (right) ↗ (upperright) ↘ (lowerright) at each point.
Followup: optimize 2d DP to 1d DP of linear extra space.
Followup: what if some cells are blocked
System Design
Availability test/debug on distributed system. Discussed and drafted about failover, replication, NoSQL etc.
Interviewer seemed to be expecting more but time ran out. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  2of 2 votes
AnswersConvert a string with digits into a literal representation of the number like: 1001 to one thousand one
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Uber Software Engineer Algorithm  2of 2 votes
AnswerGoogle
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1st round
Given a box with N balls in it, each ball having a weight, randomly choose pick one out base on the weight.
Input ball1> 5kg, ball2 > 10kg and ball3 > 35kg,
then Prob(ball1 chosen) = 10%, Prob(ball2) = 20%,
Prob(ball3) = 70% ;
Followup:
Select a ball randomly based on weights. Once a ball is chosen, remove it. Next time select from the remaining balls. Go on until there is nothing left in the box. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  3of 3 votes
AnswersFind whether string S is periodic.
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Periodic indicates S = nP.
e.g.
S = "ababab", then n = 3, and P = "ab"
S = "xxxxxx", then n = 1, and P = "x"
S = "aabbaaabba", then n = 2, and P = "aabba"
follow up:
Given string S, find out the P (repetitive pattern) of S. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  2of 2 votes
AnswersDesign a hit counter which counts the number of hits received in the past 5 minutes.
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Each function accepts a timestamp parameter (in seconds granularity) and you may assume that calls are being made to the system in chronological order (ie, the timestamp is monotonically increasing). You may assume that the earliest timestamp starts at 1.
Example:
HitCounter counter = new HitCounter();
// hit at timestamp 1.
counter.hit(1);
// hit at timestamp 2.
counter.hit(2);
// hit at timestamp 3.
counter.hit(3);
// get hits at timestamp 4, should return 3.
counter.getHits(4);
// hit at timestamp 300.
counter.hit(300);
// get hits at timestamp 300, should return 4.
counter.getHits(300);
// get hits at timestamp 301, should return 3.
counter.getHits(301);
Followup:
Due to latency, several hits arrive roughly at the same time and the order of timestamps is not guaranteed chronological.
Follow up 2:
What if the number of hits per second could be very large? Does your design scale? Report Duplicate  Flag  PURGE
Dropbox Software Engineer Algorithm  2of 4 votes
AnswersRound3 Google
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For N light bulbs , implement two methods
I. isOn(int i)  find if the ith bulb is on or off.
II. toggle(int i, int j)  i <= j. Switch state (switch on if it's off, turn off if it's on) of every bulb in range i to j.
All bulbs are off initially. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  3of 3 votes
Answers/**
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* Google
* Given a list of nonnegative numbers and a target integer k,
* write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.
**/ Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  2of 2 votes
AnswerTwitter
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Create a simple stack which takes a list of elements.
Each element contains a stack operator (push, pop, inc) and a value to either push/pop or two values, n and m,
which increment the bottom n values by m.
Then print the topmost value of the stack after every operation. If the stack is empty, print "empty" Report Duplicate  Flag  PURGE
Twitter Software Engineer Algorithm  3of 3 votes
AnswersDropBox Dec 2017
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Congrats to Brian landing @DropBox
Dropbox always has the most responsive HR and gives review on the interview within a week. The canteen is also one of the best in Bay Area.
Phone：
LC: game of life
What if the board is huge?
Store in disk
with bitmap
Load into memory, process and write to disk row by row
Visit AONECODE.COM for private lessons by FB, Google and Uber engineers
Our ONE TO ONE class offers
SYSTEM DESIGN Courses (highly recommended for candidates of FB, LinkedIn, Amazon, Google & Uber etc.)
ALGORITHMS (conquer DP, Greedy, Graph, Advanced Algos & Clean Coding),
latest interview questions sorted by companies,
mock interviews.
Members get hired from G, U, FB, Amazon, LinkedIn, MS and other toptier companies after weeks of training.
Email us aonecoding@gmail.com with any questions. Thanks!
Onsite：
Round 1：
Given an array of byte and a file name, find if the array of byte is in file.
Solution: Rolling hash
Round 2：
Given an Iterator of some photo with IDs, find the top K most hit photo IDs.
Follow up: What if the input is from a stream? When iterator reaches the end, moments later new hits can be added to the iterator. Modify code for this scenario.
Lunch was great.
Then came a demo round. Discussed Dropbox Paper Report Duplicate  Flag  PURGE
Dropbox Software Engineer  3of 3 votes
AnswersCongrats on aonecode.com member V.S. on the offer from Microsoft and thanks for sharing with us the experience.
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Coding Question 1  Find all the paths between two nodes
Coding question 2 : Max sum in adjacent sub array
Design Question  Design a ticketing System
Design Question 2  Design a system which allows multiple agents to read different data from same tables. Latency should be low. Algorithm should rank agents through some logic and assigned work according to that so that each agents are reading different set of rows from same table. Scale it for 20 million active agents .
Follow up  If Data Sharding is allowed, what will be the Shard Id and how the partition will look like? How your system will respond if there are agents which are also writing at same time. Consistency should be given high preference over availability.
Looking for coaching on interview prep?
Visit AONECODE.COM for ONETOONE private lessons by FB, Google and Uber engineers!
Customized course covers
System Design (for candidates of FB, LinkedIn, AMZ, Google & Uber etc)
Algorithms (DP, Greedy, Graph etc. every aspect you need in a coding interview & Clean Coding)
Interview questions sorted by companies
Mock Interviews
Our members got into G, U, FB, Amazon, LinkedIn, MS and other toptier companies after weeks of training.
Feel free to email us aonecoding@gmail.com with any questions. Thanks! Report Duplicate  Flag  PURGE
Microsoft Software Engineer Algorithm  2of 4 votes
AnswerCongrats to aonecode's member F.L.
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Got offers from  Youtube(G), LinkedIn, Airbnb, Square, Wish, Blend and NextDoor!
Thanks for sharing the interview experience with us.
Youtube Interview
 Phone: Find anagrams of string A from string B (sliding window)
 Phone: Find if two frames in a screen are equal. Frames may overlap. (equal method)
Onsite:
 LC41 first missing positive
 LC499+LC505 The maze
 LC161 one edit distance
 Similar to hangman but make guesses based on a dictionary.
Assume a dictionary has words  {house, morse, jesus} and ‘morse’ is the answer. If your first guess is ‘house’, output will be ‘_o_se’, which indicates 3 letters are correct. (Here the arrangement of letters does not matter. Your guess can be ‘co’ and if answer is ‘ok’ then the output is gonna be ‘_o’ which indicates letter ‘o’ in answer. )
Try to get the answer with minimum guesses.
(Interviewer expects preprocessing the dictionary. Key: letter; Value: frequency. Begin with combinations of most frequent letters first) Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  1of 3 votes
AnswerCongrats to F.L.
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Got offers from  Youtube(G), LinkedIn, Airbnb, Square, Wish, Blend and NextDoor!
Thanks for sharing the interview experience with us.
LinkedIn
Phone:
 Questions from LC tagged LinkedIn.
Onsite:
 Get sqrt(x). Output a floored integer if result is not a perfect square. sqrt(18) = 4
 Implement BST, insert, delete, search.
 Design a dashboard for service logs stats (sort and aggregate). Scale from 1 to more machines. Discuss async and realtime as different scenarios. Report Duplicate  Flag  PURGE
Linkedin Software Engineer Algorithm  2of 2 votes
AnswerFacebook
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Phone: LC304 & longest arithmetic sequence. Return the sequence. Report Duplicate  Flag  PURGE
Facebook Software Engineer Algorithm  4of 4 votes
AnswerWish Interview
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Phone: Two sum, Three sum, N sum(recursion)
Onsite:
Implement merge sort (recursion&iteration)
Merge two sorted arrays: one of length m+n, the other n; store the result in the longer array
Given a number print diamond:
Given 1
Pirnt 1
Given 3
Print
1
121
1
Given 5
Print
1
121
12321
121
1
 Rank N people in a game. There may be a tie among participants. How many possible ways of ranking there is.
 Design: Define a bot as an IP that hits the web app over M times in the past T seconds (not necessarily hits on the same page. Also take into account different API calls.) How to design a bot detector layer and where to place it in the system. Report Duplicate  Flag  PURGE
Wish Solutions Engineer Algorithm  2of 2 votes
AnswerAirbnb Online Assessment Paginate List
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5
13
1,28,310.6,SF
4,5,204.1,SF
20,7,203.2,Oakland
6,8,202.2,SF
6,10,199.1,SF
1,16,190.4,SF
6,29,185.2,SF
7,20,180.1,SF
6,21,162.1,SF
2,18,161.2,SF
2,30,149.1,SF
3,76,146.2,SF
2,14,141.1,San Jose
Here is a sample input. It’s a list generated by user search.
(1,28,100.3,Paris) corresponds to (Host ID, List ID, Points, City).
5 in the first row tells each page at most keeps 5 records.
13 in the second row is the number of records in the list.
Please paginate the list for Airbnb by requirement:
1. When possible, two records with same host ID shouldn’t be in a page.
2. But if no more records with nonrepetitive host ID can be found, fill up the page with the given input order (ordered by Points).
Expected output:
1,28,310.6,SF
4,5,204.1,SF
20,7,203.2,Oakland
6,8,202.2,SF
7,20,180.1,SF
6,10,199.1,SF
1,16,190.4,SF
2,18,161.2,SF
3,76,146.2,SF
6,29,185.2,SF  6 repeats in page bec no more unique host ID available
6,21,162.1,SF
2,30,149.1,SF
2,14,141.1,San Jose Report Duplicate  Flag  PURGE
Airbnb Software Engineer Algorithm  1of 1 vote
Answerscreate a class of integer collection,
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3 APIs:
append(int x),
get(int idx),
add_to_all(int x)，
in O(1) time。
Followup:
implement
multiply_to_all(int x)
e.g.
insert(1)
insert(2)
add_to_all(5)
insert(3)
get(0) > returns 6
get(2) > return 3
multiply_to_all(10)
insert(4)
get(1) > returns 70
get(2) > returns 30
get(3) > returns 4 Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  0of 0 votes
AnswersLonely Pixel
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Given an N x M image with black pixels and white pixels, if a pixel is the only one in its color throughout its entire row and column, then it is a lonely pixel. Find the number of lonely pixels in black from the image. (O(NM)) Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  1of 1 vote
AnswersEmployees Per Department
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Twitter Interview Online Test SQL
A company uses 2 data tables, Employee and Department, to store data about its employees and departments.
Table Name: Employee
Attributes:
ID Integer,
NAME String,
SALARY Integer,
DEPT_ID Integer
Table Name: Department
Attributes:
DEPT_ID Integer,
Name String,
LOCATION String
View sample tables:
https://s3uswest2.amazonaws.com/aonecode/techblog/50cfcdd1d61f1bd6002cf4d3b4a61debmin.jpeg
Write a query to print the respective Department Name and number of employees for all departments in the Department table (even unstaffed ones).
Sort your result in descending order of employees per department; if two or more departments have the same number of employees, then sort those departments alphabetically by Department Name. Report Duplicate  Flag  PURGE
Twitter Software Engineer SQL  0of 0 votes
AnswersI. Closest K nodes to a target in BST? (Do it in O(n)?)
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II. Nested List sum? Report Duplicate  Flag  PURGE
Linkedin Software Engineer Algorithm  0of 0 votes
AnswerSolve the 24 Game
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Twitter Software Engineer Algorithm  3of 3 votes
AnswersRound4
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Starting from num = 0, add 2^i (where i can be any nonnegative integer) to num until num == N. Print all paths of how num turns from 0 to N.
For example if N = 4,
Paths printed are [0,1,2,3,4], [0,1,2,4], [0,1,3,4], [0,2,4], [0,2,3,4], [0,4].
[0,2,4] is made from 0 + 2^1 + 2^1. [0,1,3,4] from 0 + 2^0 + 2^1 + 2^0 Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  2of 2 votes
AnswersRound1
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Find if two people in a family tree are bloodrelated.
Round2
Given some nodes in a singly linked list, how many groups of consecutively connected nodes there is.
For linked list
0>1>2>3>4>5>6,
given nodes 1, 3, 5, 6
there are 3 groups [1], [3], and [5, 6]. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  1of 1 vote
AnswersRound 5:
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Given a set of synonyms such as (fast, quick), (fast, speedy), (learn, study), decides if two sentences were synonymous.
(The sentences were structurally the same and has the same number of words in them.
The synonymous relation [fast ~ quick] and [fast ~ speedy] does not necessarily mean [quick ~ speedy].)
Followup:
If the synonymous relation passes down so that [fast ~ quick] and [fast ~ speedy] implies [quick ~ speedy], decide if two sentences were synonymous. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  1of 1 vote
AnswersRound 4:
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Implement a class Employment with these 3 methods: assignManager(p1, p2): assign p1 as p2's manager. beColleague(p1, p2): make p1 and p2 peer colleagues. isManager((p1, p2): decide if p1 is the manager of p2. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  2of 2 votes
AnswersRound 3
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Given a matrix of 0s and 1s where 0 is wall and 1 is pathway, print the shortest path from the first row to the last row.
Can walk to the left, top, right, bottom at any given spot.
Followup:
If every pathway takes a cost (positive integer) to get through, print the minimum cost path from the first row to the last row. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  1of 1 vote
AnswersGoogle onsite June
 aonecoding in United States
Round 1
Leetcode 10
Round 2
Select a random point uniformly within a rectangle, (The side of rectangle is parallel to the x/ y grid).
Followup: Given multiple nonoverlapped rectangles on the 2D grid, uniformly select a random point from the rectangles. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  2of 2 votes
AnswersPhone Interview Amazon, Seattle
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I. Get the sum of all prime numbers up to N. primeSum(N).
Followup: If primeSum(N) is frequently called, how to optimize it.
II. OODesign Parking Lot Report Duplicate  Flag  PURGE
Amazon Software Engineer Algorithm  2of 2 votes
AnswersApple Map Team
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1. Given an array A and some queries, query(i, j) returns the result of Ai*...*Aj, in other words the multiplication from Ai to Aj.
The numbers in A are nonnegative.
Implement query(i, j).
2. Flatten nested linked list
3. POI search design
4. LC238 & LC279 Report Duplicate  Flag  PURGE
Apple Software Engineer Algorithm  1of 1 vote
AnswersAirbnb Interview
 aonecoding in United States
Min cost of flight from start to end if allowed at most k transfers.
Given all the flights in a string:
A>B,100,
B>C,100,
A>C,500,
If k = 1，from A to C the best route is A>B>C at the cost of 200. Report Duplicate  Flag  PURGE
Airbnb Algorithm  3of 3 votes
AnswersApple phone interview
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Given an API to find all IPv4 addresses in a log file, find all IPs that occurred only once.
Followup: What if the log comes from a data stream.
Followup: If the machine has 4GB RAM, is there going to be a problem? Report Duplicate  Flag  PURGE
Apple Backend Developer Algorithm  2of 2 votes
Answers4/5 Round at Uber
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Coding: Given a 2D array of either '\' or '/', find out how many pieces this rectangle is divided into graphically.
For a 2X2 matrix with
/\
\/
The matrix split into 5 pieces  the diamond in middle and the four corners. Return 5 as the answer.
5/5 Round at Uber
Design Excel. Report Duplicate  Flag  PURGE
Uber Software Engineer Algorithm  1of 1 vote
Answers2/5 Round at Uber
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Bar raiser  Behavioral questions. Coding: Find if a set of meetings overlap. Meeting has a starttime and an endtime with accuracy to minute. All meetings take place in the same day. Do this in O(n) time.
3/5 Round at Uber
Coding: Subset sum. Followup: Optimize the solution. Report Duplicate  Flag  PURGE
Uber Software Engineer Algorithm  2of 2 votes
AnswerUber
 aonecoding in United States
1. Mirror Binary Tree
2. String pattern matching
The matching should cover the entire input string (not partial).
The function prototype should be:
bool isMatch(String str, String pattern)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa","a{1,3}") → true
isMatch("aaa","a{1,3}") → false
isMatch("ab","a{1,3}b{1,3}") → true
isMatch("abc","a{1,3}b{1,3}c") → true
isMatch("abbc","a{1,3}b{1,2}c") → false
isMatch("acbac","a{1,3}b{1,3}c") → false
isMatch("abcc","a{1,3}b{1,3}cc{1,3}") → true
In pattern string, a char followed by {lower, upper} means that the char occur lower to upper(exclusive) times. e.g. a{1, 3} > a occurs 1 or 2 times. Report Duplicate  Flag  PURGE
Uber Software Engineer Algorithm  3of 3 votes
Answers3.1 design: design fb inbox search —> just focus on the post
 aonecoding in United States
4.1 binary tree to circular double linked list.
4.2 two arrays, find the common elements of two sorted array. if one array is small, the other is very big. Report Duplicate  Flag  PURGE
Facebook Software Engineer Algorithm  2of 2 votes
Answers2.1 career discussion
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2.2 divide two numbers with no / or % Report Duplicate  Flag  PURGE
Facebook Software Engineer Algorithm  3of 3 votes
Answers1 year exp. Interviewed at Cambridge, MA
 aonecoding in United States
Round1
LC304. Followup: given a char stream instead a string as the input, get the longest substring with at most K distinct characters.
Round2
Find out the area of a number of squares on a plane, an advanced version of LC223.
Had no clue on that problem at all so the interviewer kindly gave another one LC305.
Round3
Similar to LC393 but the interviewer made a slightly different rule for encoding.
Followup: decode with utf16. It took quite a while for me to understand the rules.
Round4
Card game rule: the hand is drawn from a pack of cards (no jokers).
Play cards ONLY when they are
1. 3 of a kind ('AAA' ) or 4 of a kind('AAAA’).
2. a straight flush of 3 or more cards('JQK' or 'A23456...' in the same suit).
Find out whether the player is able to play the whole hand given.
e.g. [Spade A, Spade K, Spade Q, Diamond Q, Heart Q] return false.
[Spade A, Spade K, Spade Q, Diamond Q, Heart Q, Club Q] return true. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  3of 3 votes
AnswersGoogle fulltime phD candidate w/ work experience.
 aonecoding in United States
Q1. On a 1 meter walkroad, randomly generate rain. The raindrop is 1 centimeter wide.
Simulate how raindrops cover the 1 meter [0~1] road. How many drops does it take to fully cover the 1meter?
Q2. Find out the maximum number of isosceles triangles you can make from a plane of points(cannot reuse points)
Q3.Longest holiday  Every city has a different days of holidays every week. You may only travel to another city at the weekends. What is the max days of holiday you can get this year.
Q4.
Design merchandising product data storage service Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  5of 5 votes
AnswersGoogle Onsite in May
 aonecoding in United States
Create a class with a collection of integers.
Enable 3 APIs:
void append(int x),
int get(int idx),
void add_to_all(int x)，//add x to all numbers in collection
These methods should run in O(1) time.
Followup
In addition, implement
void multiply_to_all(int x)
The same required to run in O(1) time Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  1of 1 vote
AnswersImplement circular buffer with read & write functions
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Facebook Software Engineer Data Structures  1of 1 vote
AnswersCoding III
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Implement int divide(int a, int b) without division or mod operation.
## Round IV
Behavioral Questions + Project Walk Through + Coding (Validate BST)
## System Design V
Design memcache to enable read, write and delete (single server, nondistributed infrastructure).
Which data structure to go with?
Eviction rules?
How to minimize segmentation?
How to handle concurrency?
## Extra
After two weeks they called me to an extra round of system design.
How to store social graphs?
How to handle concurrent read/write requests(read heavy) on one server. Report Duplicate  Flag  PURGE
Facebook Software Engineer  1of 1 vote
AnswersLast Monday phone interview of G.
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Given a vector/list of doubly linked list pointers (a pointer is the directed linkage of two nodes), count how many independent blocks of linked lists there are for the pointers given. Report Duplicate  Flag  PURGE
Google SDE1 Algorithm  2of 2 votes
AnswersApple Onsite at Cupertino
 aonecoding in United States
Team Data Warehousing
Questions on Hadoop, Hive and Spark
I. Given a table with 1B of user ID and product IDs that the users bought, and another table with product ID mapped with product name. We are trying to find the paired products that are often purchased together by the same user, such as wine and bottle opener, chips and beer … How to find the top 100 of these coexisted pairs of products. If going with hadoop, where is the bottleneck and how to optimize?
II. Someone put distribute Random()*ID in a Hive script to prevent data skew. What would be the problem here? Report Duplicate  Flag  PURGE
Apple SDE3 design  1of 1 vote
AnswersApple Onsite at Cupertino
 aonecoding in United States
Team Data Warehousing
III. Given three letters ABC, where AB>C, AC>B, BC>A (sequence doesn’t matter). Get the length of the path to convert from a given string to a single character.
For example, “ABACB” goes to “ACCB” (based on AB >C, convert s[1] and s[2] to C)
“ACCB” goes to “BCB” (based on AC>B)
“BCB” goes to “AB”
“AB” goes to “C”
So it takes 4 steps to change the given string into a single character.
If a given string cannot be resized to 1 character, such as “AAA” or "ABACABB", return 1. Report Duplicate  Flag  PURGE
Apple SDE3 Algorithm  1of 1 vote
AnswersApple Onsite at Cupertino
 aonecoding in United States
Team Data Warehousing
There were 6.5 rounds in total, that 0.5 on package negotiation with the HR and the remaining rounds with 2 managers and 4 engineers.
Only three pure coding questions were asked.
I. Use a stack to sort given data.
II. Given an array with positive integers only, find the MIN integer that is missing from the array. Report Duplicate  Flag  PURGE
Apple SDE3 Algorithm  1of 3 votes
AnswersAmazon SDE 2 Onsite (4 of 4 Rounds)
 aonecoding in United States
Assume that there is an ebook application. For every book the sharable part of the book content cannot exceed 10% of the whole book. Design a module to decide whether the current part of content is sharable.
The description given is vague. I had to push him with questions to give the details.
At first I thought the problem was about strStr. But then the interviewer said that even if there are two paragraphs of the book content with the exact same texts, as long as they are not in the same place, they would be considered different content.
I then realized it’s a question about merging segments  have a helper to find each pair of start and end point of the input content (given multiple separated paragraphs). Then merge the intervals and see if they combined exceed 10% of the entire book.
The interviewer approved my solution and ask me to code it.
Overall I feel like that the Amazon SDE II Interview doesn’t focus on just algorithm. It’s more about problem solving in practice and then implement the only core function on whiteboard. Report Duplicate  Flag  PURGE
Amazon Software Engineer  1of 1 vote
AnswersVMWare Standard Online Screen
 aonecoding in United States
3rd Question Given an array of strings and a long description about the formatting of IPv6 and IPv4 (it took me more than 5 minutes to read the description). Write a function to find if a string is IPv4 or IPv6 address or neither.
4th Question Given an integer array, whenever a duplicate number is found, you may increment it (++). Find the minimum sum of the numbers in the array by keep incrementing the dups until all the numbers are unique. Report Duplicate  Flag  PURGE
VMWare Inc Software Engineer Algorithm  2of 2 votes
AnswerVMWare Standard Online Screen
 aonecoding in United States
The Online Assessment was called something like Life Cycle Challengeqpan.
There are 4 questions in total given 60 minutes. The problem description was unexpectedly long that it takes 5 minutes just to read a question.
1st Question Design a function to create BST. Given an integer array, insert the integers into the binary search tree and print all the steps taken.
2nd Question Given an integer, print the index of all the positions in which the binary bit is 1 in order. Report Duplicate  Flag  PURGE
VMWare Inc Software Engineer Algorithm  7of 7 votes
AnswersFacebook Senior Engineer Onsite 2017
 aonecoding in United States
1st Round
Question 1: Binary tree to doubly linked list.
Question 2: Read 4 (Given the read4 API, read an entire file)
2nd Round
Culture fit. No coding.
3rd Round
Question: System Design POI (Point of Interest. Given a point, find things within a radius).
Lunch
4th Round
Question 1: Decode way
Question 2: Random max index
5th Round
Question: System design + componentwise design on download manager Report Duplicate  Flag  PURGE
Facebook Software Engineer Algorithm  6of 6 votes
Answers5th Round
 aonecoding in United States
Openended question What happens when you type a url in the browser and hit enter?
Second question Given an array of integers, print all the numbers that meet the following requirement  when the number is greater than every number on its left and smaller than every number on the right. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  2of 4 votes
Answerinterviewed by senior engineer
 aonecoding in United States
Question Given two strings s1 and s2, combine the characters in the strings and maintain the sequence of characters
Followup If s1 has a length of m and s2 has a length of n, how many ways the strings could be merged. Figure out the formula F(m, n) = ? Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  0of 2 votes
AnswersHow to randomly select a number in an array?
 aonecoding in United States
array: [15, 2, 4, 5, 1, 2, 0]
Followup:
Given a second array freq where freq[i] represents the occurrence of the ith number in array, how to randomly select a number in array based on the frequency.
Extra requirement:
Could you complete the selection in a singlepass(go through each array only once)? Report Duplicate  Flag  PURGE
Linkedin Software Engineer Algorithm  1of 5 votes
AnswersIn school a student gets rewarded if he has an attendance record without being absent for more than once or being late for 3 times continuously.
 aonecoding in United States
Given a student's attendance record represented by a string with 3 possible characters 'L'(Late), 'A'(Absent), 'O' (On time),
check whether the student qualifies for the reward.
e.g.
@INPUT (String) "OLLAOOOLLO"
@RETURN (Boolean) False
The student does not qualify for reward because "LLA" means he was late for 3 times in a row.
@INPUT (String) "OLLOAOLLO"
@RETURN (Boolean) True
Followup:
If known the length of the attendance string is n, how many possible ways there is to attend school and make sure you get the reward. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  0of 0 votes
AnswersIn school a student gets rewarded if he has an attendance record without being absent for more than once or being late for 3 times continuously.
 aonecoding in United States
Given a student's attendance record represented by a string with 3 possible characters 'L'(Late), 'A'(Absent), 'O' (On time), check whether the student qualifies for the reward.
e.g.
@INPUT (String) "OLLAOOOLLO"
@RETURN (Boolean) False
The student does not qualify for reward because "LLA" means he was late for 3 times in a row.
@INPUT (String) "OLLOAOLLO"
@RETURN (Boolean) True
Followup:
If known the length of the attendance string is n, how many possible ways there is to attend school and make sure the student gets the reward. Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  0of 6 votes
AnswersCreate an iterator class that stores a list of the builtin Iterators.
 aonecoding in United States
Implement the next() and hasNext() methods in a Round Robin pattern (pops next element in a circle).
Example:
Given a list [iterator1,iterator2, iterator3...]
when calling RoundIterator.next()
pops iterator1.next if iterator1.hasNext() is true
when calling RoundIterator.next()
pops iterator2.next() if iterator2.hasNext() is true
when calling RoundIterator.next()
pops iterator3.next if iterator3.hasNext() is true
...
when calling RoundIterator.next()
pops iterator1.next if iterator1.hasNext() is true
when calling RoundIterator.next()
pops iterator2.next if iterator2.hasNext() is true
when calling RoundIterator.next()
pops iterator3.next if iterator3.hasNext() is true
...
until there is no more element in any of the iterators Report Duplicate  Flag  PURGE
Google Algorithm  1of 9 votes
AnswersQ: Weighted meeting room
Given a series of meetings, how to schedule them. Cannot attend more than a meeting at the same time. Goal is to find maximum weight subset of mutually nonoverlap meetings.class Meeting: def __init__(self): self.startTime self.endTime self.weight
@concernedCoder
 aonecoding in United States
When you claim the questions as fake, provide evidence. These are no doubt questions asked in the coding interviews of the best companies and they definitely help interviewees to prepare for the interview.
Why do you have a problem with this? Report Duplicate  Flag  PURGE
Facebook Software Engineer Algorithm  4of 4 votes
Answers# There's a room with a TV and people are coming in and out to watch it. The TV is on only when there's at least a person in the room.
 aonecoding in United States
# For each person that comes in, we record the start and end time. We want to know for how long the TV has been on. In other words:
# Given a list of arrays of time intervals, write a function that calculates the total amount of time covered by the intervals.
# For example:
# input = [(1,4), (2,3)]
# > 3
# input = [(4,6), (1,2)]
# > 3
# input = {{1,4}, {6,8}, {2,4}, {7,9}, {10, 15}}
# > 11 Report Duplicate  Flag  PURGE
Facebook Software Engineer Algorithm  6of 6 votes
Answers/**
 aonecoding in United States
Given many coins of 3 different face values, print the combination sums of the coins up to 1000. Must be printed in order.
eg: coins(10, 15, 55)
print:
10
15
20
25
30
.
.
.
1000
*/ Report Duplicate  Flag  PURGE
Facebook Software Developer Algorithm  2of 4 votes
AnswersFind all comments in the Java (it could be Python or any other language of your choice) codes that’s parsed in as a string.
 aonecoding in United States
You may assume the codes given is valid.
Input is a single string, e.g.
String codes =
“/* file created by aonecode.com\\n” +
“ welcome to the tech blog*/ \\n” +
“//main method\\n” +
“public static void main(String[] args) { \\n“ +
“ System.out.println(“//welcome”); //output\\n” +
“}”
Output is a list of strings
List<String> ret =
[
“ file created by anecode.com\n welcome to the tech blog”,
“main method”,
“output”
] Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm  5of 5 votes
AnswersQ: If you were given a series of equations e.g. [A = B, B = D, C = D, F = G, E = H, H = C]
 aonecoding in United States
and then another series [A != C, D != H, ..., F != A ]
Check whether the equations combined is valid.
For the example given, your program should return 'invalid', because the first series implies that A = C, which contradicts the statement A != C in the second series. Report Duplicate  Flag  PURGE
Amazon Software Engineer Algorithm  3of 3 votes
AnswersGiven a list of system packages, some packages cannot be installed until the other packages are installed. Provide a valid sequence to install all of the packages.
 aonecoding in United States
e.g.
a relies on b
b relies on c
then a valid sequence is [c, b, a] Report Duplicate  Flag  PURGE
Uber Software Engineer Algorithm  3of 5 votes
AnswersQ: Find the absolute paths to all directories with image files, given a file system that looks like this. The subdirectory is one indent over.
 aonecoding in United States/usr /local profile.jpg /bin config.txt dest.png /rbin image.gif /sys /re /tmp pic.jpg ..... ……
 Report Duplicate  Flag  PURGE
Google Software Engineer Algorithm
 0 Answers Coding Interview Mentoring
Looking for interview questions sharing and mentors? Visit A++ Coding Bootcamp at aonecode.com (select english at the top right corner).
 aonecoding January 15, 2017
We provide ONE TO ONE courses that cover everything in an interview from the latest interview questions, coding, algorithms, system design to mock interviews. All classes are given by experienced engineers/interviewers from FB, Google and Uber. Help you close the gap between school and work. Our students got offers from G, U, FB, Amz, Yahoo and other top companies after a few weeks of training.
Welcome to email us with any questions. Flag  PURGE  0 Answers Coding Interview Mentoring
Looking for interview questions sharing and mentors? Visit A++ Coding Bootcamp at aonecode.com (select english at the top right corner).
 aonecoding January 15, 2017
We provide ONE TO ONE courses that cover everything in an interview from the latest interview questions, coding, algorithms, system design to mock interviews. All classes are given by experienced engineers/interviewers from FB, Google and Uber. Help you close the gap between school and work. Our students got offers from G, U, FB, Amz, Yahoo and other top companies after a few weeks of training.
Welcome to email us with any questions. Flag  PURGE
Looking for interview experience sharing and coaching?
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SYSTEM DESIGN
ALGORITHMS (conquer DP, Greedy, Graph, Advanced Algorithms, Clean Coding),
latest interview questions sorted by companies,
mock interviews.
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The optimal solution is a little tricky. Can you prove it?
Optimal Solution:
public int shortestNonSeq(int[] a, int k) {
int result = 0;
Set<Integer> flag = new HashSet();
for (int i = 0; i < a.length; i++) {
if (!flag.contains(a[i])) {
flag.add(a[i]);
if (flag.size() == k) {
result++;
flag = new HashSet<>();
}
}
}
return result + 1;
}

aonecoding
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SYSTEM DESIGN
ALGORITHMS (conquer DP, Greedy, Graph, Advanced Algorithms, Clean Coding),
latest interview questions sorted by companies,
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SOLUTION:
public int connectedComponents(int n, int[][] edges) {
int[] uf = new int[n];
for(int i = 0; i < n; i++) {
uf[i] = i;
}
for(int[] e : edges) {
int x = e[0];
int y = e[1];
while (uf[x] != x){
x = uf[x];
}
while (uf[y] != y){
y = uf[y];
}
if(x != y) {
uf[x] = y;
n;
}
}
return n;
}

aonecoding
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SOLUTION:
int assignBalls(int m, int n) {
if (m == 0  n == 1) {
return 1;
}
if (n > m) {
return assignBalls(m, m);
} else {
return assignBalls(m, n  1) + assignBalls(m  n, n);
}
}

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SOLUTION:
int smallestFactors(int n) {
if (n < 10) {
return n;
}
List<Integer> list = new ArrayList<>();
for (int i = 9; i > 1; i) {
while ((n % i) == 0) {
n /= i;
list.add(i);
}
}
if (n > 10) {
return 0;
}
int result = 0;
for (int i : list) {
result = result * 10 + i;
}
return result;
}
followup: what if we need to worry about the integer overflow?
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SYSTEM DESIGN
ALGORITHMS (conquer DP, Greedy, Graph, Advanced Algorithms, Clean Coding),
latest interview questions sorted by companies,
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SOLUTION:
boolean find(String s1, String s2) {
if(s1.isEmpty()) return false;
int l1 = s1.length(), l2 = s2.length();
if(l1 >= l2) {
return kmp(s1 + s1, s2); //find if s2 is a substring of s1 + s1
} else {
for(int i = 0; i < l1; i++) {
if(match(s1, i, s2)) return true; //find if s2 is made of multiple rotated s1
}
}
return false;
}
boolean match(String s1, int j, String s2) {
int idx = 0;
while(idx < s2.length()) {
if(s2.charAt(idx) != s1.charAt(j)) {
return false;
}
idx++;
j = j % s1.length();
}
return true;
}
boolean kmp(String haystack, String needle) {
if (haystack.length() < needle.length()) {
String t = haystack;
haystack = needle;
needle = t;
}
int[] table = new int[needle.length() + 1];
for (int i = 1; i < needle.length(); i++) {
int j = table[i];
while (j > 0 && needle.charAt(i) != needle.charAt(j)) {
j = table[j];
}
table[i + 1] = (needle.charAt(i) == needle.charAt(j)) ? j + 1 : 0;
}
haystack += haystack;
int j = 0;
for (int i = 0; i < haystack.length(); i++) {
while (j > 0 && needle.charAt(j) != haystack.charAt(i)) {
j = table[j];
}
if (needle.charAt(j) == haystack.charAt(i)) {
j++;
}
if (j == needle.length()) {
return true;
}
}
return false;
}

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class RingBuffer:
def __init__(self, capacity):
self.size = capacity
self.capacity = 0
self.Buffer = [None] * capacity
self.Head = 0
self.Tail = 0
def full(self):
return self.size == self.capacity
def empty(self):
return self.capacity == 0
def add(self, obj):
self.Buffer[self.Tail] = obj
self.Tail = (self.Tail + 1) % self.size
Self.capacity += 1
def pop(self):
self.Head = (self.Head + 1) % self.size
Self.capacity = 1
return self.Buffer[self.Head  1]

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public void backButton(String[] cmds) {
Stack<String> forwardStack=new Stack();
Stack<String> backwardStack=new Stack();
String currentPage="about:blank";
for(int i = 0; i < cmds.length; i++)
{
String cmd = cmds[i];
if(cmd.equals("open"))
{
backwardStack.push(currentPage);
currentPage=cmds[++i];
forwardStack = new Stack();
}
else if(cmd.equals("back"))
{
if(!backwardStack.isEmpty()){
forwardStack.push(currentPage);
currentPage=backwardStack.pop();
}
}
else if(cmd.equals("forward"))
{
if(!forwardStack.isEmpty()){
backwardStack.push(currentPage);
currentPage=forwardStack.pop();
}
}
}
return currentPage;
}

aonecoding
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SOLUTION:
BFS approach does it
public List<Integer> leftView(TreeNode root) {
List<Integer> leftview = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<>();
if(root != null) q.add(root);
while(!q.isEmpty()) {
leftview.add(q.peek().val);
Queue<TreeNode> nextLevel = new LinkedList<>();
while(!q.isEmpty()) {
TreeNode node = q.poll();
if(node.left != null) nextLevel.add(node.left);
if(node.right != null) nextLevel.add(node.right);
}
q = nextLevel;
}
return leftview;
}

aonecoding
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SOLUTION:
ExpTree build() is not required. Solution from equals() below
from collections import defaultdict
class TreeNode():
def __init__(self, val):
self.val = val
self.left = None
self.right = None
if val != '+' and val != '':
self.isLeaf = True
else:
self.isLeaf = False
class ExpressionTree():
def __init__(self, exp):
self.root = self.build(exp)
def build(self, exp):
if not exp:
return
if len(exp) == 1:
return TreeNode(exp)
mid = (len(exp)  1) / 2
if exp[mid] != '+' and exp[mid] != '':
mid += 1
if exp[mid] == '':
exp = self.flip(exp, mid + 1)
node = TreeNode(exp[mid])
node.left = self.build(exp[:mid])
node.right = self.build(exp[mid + 1:])
return node
def flip(self, exp, idx):
ls = list(exp)
for i in range(idx, len(exp)):
if ls[i] == '+':
ls[i] = ''
elif ls[i] == '':
ls[i] = '+'
return "".join(ls)
def equals(self, other):
#print self.result()
#print other.result()
return self.result() == other.result()
def result(self):
d = self.calculate(self.root)
for key, value in d.items():
if value is 0:
del d[key]
return d
def calculate(self, node):
if not node:
return defaultdict(lambda:0)
if node.isLeaf:
d = defaultdict(lambda:0)
d[node.val] += 1
return d
left = self.calculate(node.left)
right = self.calculate(node.right)
for entry in right:
left[entry] = left[entry] + (right[entry] if node.val == '+' else right[entry])
return left
A tester
A = 'a+bcabcab'
B = 'b+a+c+a+bc+b'
print ExpressionTree(A).equals(ExpressionTree(B))

aonecoding
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SOLUTION:
BFS
from heapq import heappush, heappop
import numpy as np
def minCostPath(maze):
if len(maze) is 0 or len(maze[0]) == 0 or maze[0, 0] == 1:
return
m, n = len(maze), len(maze[0])
distances = np.zeros(shape = (m, n))
distances.fill(1)
heap = []
heappush(heap, (maze[0, 0], 0, 0))
value = maze[0, 0]
distances[0, 0] = value
neighbors = [(1, 0), (1, 0), (0, 1), (0, 1)]
while heap and distances[m  1, n  1] == 1:
dist, x, y = heappop(heap)
for neighbor in neighbors:
i, j = x + neighbor[0], y + neighbor[1]
if i >= 0 and j >= 0 and i < m and j < n and maze[i, j] != 1 and distances[i, j] == 1:
heappush(heap, (maze[i, j] + dist, i, j))
distances[i, j] = maze[i, j] + dist
print distances
if distances[m  1][n  1] == 1:
return
x, y = m  1, n  1
path = []
while x != 0 or y != 0:
path.append((x, y, distances[x, y]))
x, y = getMinParent(distances, x, y, neighbors)
path.append((0, 0, maze[0, 0]))
return path
def getMinParent(distances, x, y, neighbors):
minDist = distances[x, y]
fromx, fromy = None, None
for neighbor in neighbors:
i, j = x + neighbor[0], y + neighbor[1]
if i >= 0 and j >= 0 and i < len(distances) and j < len(distances[0]) and distances[i, j] >= 0 and distances[i, j] <= minDist:
fromx, fromy = i, j
minDist = distances[i, j]
return fromx, fromy
A tester
minCostPath(np.array([[4,2,1,1,1],
[1,1,6,7,1],
[4,1,3,0,1],
[4,7,3,10,1],
[4,8,6,1,1],
]))

aonecoding
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SOLUTION:
Q1  DFS to count the number of connected components in graph.
Followup 
What if board is too big?
Consider dividing the board and process each part separately.
Divide into blocks or row by row?
Row by Row. It's hard to deal with the border of blocks.
Union find by each row
What's the memory consumption?
Each time load 1 row of input into memory. Create a union find array for two rows (current row and previous row). Memory = 3 * row.
public class NumberOfIslands {
int rowLen;
int rowCounter;
int numberOfIslands;
Integer[] islands;
public NumberOfIslands(int rowLen) {
this.rowLen = rowLen;
islands = new Integer[(rowLen + 1) * 2];
}
public int loadRow(int[] row) {
rowCounter++;
for(int i = 1; i <= rowLen; i++) {
if(row[i  1] != 0) {
int j = i + rowLen + 1;
Integer top_root = find(i);
Integer left_root = find(j  1);
//System.out.println("idx " + i );
//System.out.println("top " + top_root + " left " + left_root);
if(top_root == null && left_root == null) { //new island found
numberOfIslands++;
islands[j] = j;
} else if(top_root == null) { //no top neighbor. join left neighbor
islands[j] = left_root;
} else if(left_root == null){ //no left neighbor. join top neighbor
if(top_root <= rowLen) { //top neighbor had no connection with new row
islands[top_root] = j;
islands[j] = j;
} else { //top neighbor's ancestor is in new row
islands[j] = top_root;
}
} else {
if(top_root == left_root) { //no union since left neighbor and top neighbor were in same island
islands[j] = left_root;
} else { //union left and top neighbors
islands[top_root] = left_root;
islands[j] = left_root;
numberOfIslands;
}
}
}
}
compression();
for(int i = 0; i <= rowLen; i++) {
int j = i + rowLen + 1;
if(islands[j] != null) {
islands[i] = islands[j] % (rowLen + 1);
islands[j] = null;
} else {
islands[i] = null;
}
}
return numberOfIslands;
}
private Integer find(int idx) {
if(islands[idx] == null) return null;
while(idx != islands[idx]) {
idx = islands[idx];
}
return idx;
}
private void compression() {
for(int i = rowLen + 1; i < islands.length; i++) {
if(islands[i] != null) {
islands[i] = find(i);
}
}
}
A tester
public static void main(String[] args) {
NumberOfIslands instance = new NumberOfIslands(10);
System.out.println("number of islands " + instance.loadRow(new int[]{0,0,1,1,0,1,0,1,0,1}) + '\n');
System.out.println("number of islands " + instance.loadRow(new int[]{1,1,1,1,1,1,0,1,1,1}) + '\n');
System.out.println("number of islands " + instance.loadRow(new int[]{0,0,0,0,0,0,0,0,0,1}) + '\n');
System.out.println("number of islands " + instance.loadRow(new int[]{1,0,0,1,0,1,0,0,0,1}) + '\n');
System.out.println("number of islands " + instance.loadRow(new int[]{1,1,1,1,0,1,0,1,0,1}) + '\n');
}

aonecoding
April 21, 2018 Looking for interview experience sharing and coaching?
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SOLUTION:
public class KClosestPoints {
public class Point {
public int x;
public int y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
}
public List<Point> findKClosest(Point[] p, int k) {
PriorityQueue<Point> pq = new PriorityQueue<>(k, new Comparator<Point>() {
@Override
public int compare(Point a, Point b) {
return (b.x * b.x + b.y * b.y)  (a.x * a.x + a.y * a.y);
}
});
for (int i = 0; i < p.length; i++) {
if (i < k)
pq.offer(p[i]);
else {
Point temp = pq.peek();
if ((p[i].x * p[i].x + p[i].y * p[i].y)  (temp.x * temp.x + temp.y * temp.y) < 0) {
pq.poll();
pq.offer(p[i]);
}
}
}
List<Point> x = new ArrayList<>();
while (!pq.isEmpty())
x.add(pq.poll());
return x;
}
}

aonecoding
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Solution: O(n)
first forloop removes all invalid ')'. Second forloop removes all invalid '('
public static String balanceParenthesis(String s) {
StringBuilder str = new StringBuilder(s);
int left = 0;
for(int i = 0; i < str.length(); i++) {
if(str.charAt(i) == '(') {
left++;
} else if(str.charAt(i) == ')') {
if(left > 0) {
left;
} else {
str.deleteCharAt(i);
}
}
}
int right = 0;
for(int i = str.length()  1; i >= 0; i) {
if(str.charAt(i) == ')') {
right++;
} else if(str.charAt(i) == '('){
if(right > 0) right;
else {
str.deleteCharAt(i);
}
}
}
return str.toString();
}

aonecoding
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SOLUTION:
Seems like there are 3 cases:
1. s1 and s2 share no mutual characters. Then there is no way to further minimize the distance.
2. s1 and s2 share a pair of characters on crossed positions, such as 'b' and 'c' in "aabaaac" and "aacaaab". Then swap 'b' and 'c'.
3. s1 and s2 has no crossed pairs but has the same character on different positions. Then swap to align the character.
//@return the pair of indices in s2 to swap with.
public int[] Swap(String s1, String s2) {
if(s1.length() > s2.length()) {
return Swap(s2, s1);
}
int swap_idx1 = 1, swap_idx2 = 1;
for(int i = 0; i < s1.length(); i++) {
if(s1.charAt(i) != s2.charAt(i)) {
for (int j = i + 1; j < s1.length(); j++) {
if(s1.charAt(i) == s2.charAt(j) && s1.charAt(j) == s2.charAt(i)) {
return new int[]{i, j}; //case 2 found
} else if(s1.charAt(i) == s2.charAt(j) && s1.charAt(j) != s2.charAt(j)) {
swap_idx1 = i;
swap_idx2 = j;
}
}
for(int j = s1.length(); j < s2.length(); j++) {
if(s1.charAt(i) == s2.charAt(j)) {
swap_idx1 = i;
swap_idx2 = j;
}
}
}
}
if(swap_idx1 == 1) return null; //case1
else return new int[]{swap_idx1, swap_idx2};
}

aonecoding
April 01, 2018 Looking for coaching on interview preparation?
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SOLUTION:
import java.util.List;
import java.util.Map;
import java.util.HashMap;
class TreeNode {
int val;
List<TreeNode> kids;
}
public class HouseRobTree {
public int houseRob(TreeNode root) {
Map<TreeNode, Integer> money_in_tree = new HashMap<>();
return houseRobHelper(root, money_in_tree);
}
private int houseRobHelper(TreeNode root, Map<TreeNode, Integer> money_in_tree) {
if(root == null) return 0;
if(root.kids == null  root.kids.isEmpty()) return root.val;
if(money_in_tree.containsKey(root)) return money_in_tree.get(root);
int max = 0;
for(TreeNode kid: root.kids) {
max = Math.max(max, houseRobHelper(kid, money_in_tree));
if(kid.kids != null) {
for(TreeNode grandKid: kid.kids) {
max = Math.max(max, houseRobHelper(grandKid, money_in_tree) + root.val);
}
}
}
money_in_tree.put(root, max);
return max;
}
}

aonecoding
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SOLUTION
//O(N) time and space for processing the board and lamps
//O(1) for finding if a cell is illuminated
public class GridIllumination {
int N; //board size
Set<Integer> illuminated_x = new HashSet<>();
Set<Integer> illuminated_y = new HashSet<>();
Set<Integer> illuminated_diag0 = new HashSet();
Set<Integer> illuminated_diag1 = new HashSet();
//@param lamps  a list of (x,y) locations of lamps
public GridIllumination(int N, int[][] lamps) {
this.N = N;
for(int[] lamp: lamps) { //this lamp illuminates 4 lines of cells
illuminated_x.add(lamp[0]); //the entire column
illuminated_y.add(lamp[1]); //the entire row
illuminated_diag0.add(lamp[1]  lamp[0]); //diagonal line with a slope of 1
illuminated_diag1.add(lamp[0] + lamp[1]); //diagonal lines with a slope of 1
}
}
public boolean is_illuminated(int x, int y) {
if(illuminated_x.contains(x) 
illuminated_y.contains(y) 
illuminated_diag0.contains(y  x) 
illuminated_diag1.contains(x + y)) {
return true;
}
return false;
}
}

aonecoding
March 21, 2018 Looking for interview experience sharing and coaching?
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public class MovingAvg {
int[] q; // a circular queue of size N
int head; //queue head
int tail; //queue tail
int size; // queue size
int sum;
public MovingAvg(int N) {
q = new int[N];
}
//@param num  the next number from data stream
//@return  new average with num included and expired number excluded
public double getAverage(int num) {
double avg = 0;
sum += num;
if(size == q.length) {
sum = q[head];
head = (head + 1) % q.length;
} else {
size++;
}
q[tail] = num;
tail = (tail + 1) % q.length;
return 1.0 * sum / size;
}
}

aonecoding
March 17, 2018 Looking for interview experience sharing and coaching?
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public class MatrixPuzzle {
int column;
int row;
public MatrixPuzzle(int length, int width) {
this.column = length;
this.row = width;
}
//The dp formula will be M[i,j] = M[i  1, j  1] + M[i  1, j] + M[i  1, j + 1]
//Because one could only land on current cell from the 3 cells in the upperleft, left and lowerleft.
//To make the space consumption 1d, cache the numbers in one column of the matrix at a time.
//Followup 2: Just reset pathcounts for blocked cells to 0
public int numberOfPaths() {
int[] paths = new int[row];
paths[row  1] = 1; //start from bottomleft corner
for(int col = 1; col < column; col++) {
int upper_left_value = 0;
for(int r = 0; r < row; r++) {
int left_value = paths[r];
paths[r] += upper_left_value + (r == row  1 ? 0 : paths[r + 1]);
upper_left_value = left_value;
}
}
return paths[row  1]; //exit from bottomright
}
}

aonecoding
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//Extra space O(1) Runtime O(nlogn)
List<Integer> getFibonacciNumber(int[] nums) {
List<Integer> fibonacciNumbers = new ArrayList<>();
Arrays.sort(nums);
int fib1 = 1, fib2 = 1;
for(int i = 0; i < nums.length;) {
if(nums[i] < fib2) {
i++;
} else if(nums[i] == fib2) {
fibonacciNumbers.add(nums[i]);
i++;
} else {
int fib3 = fib1 + fib2;
fib1 = fib2;
fib2 = fib3;
}
}
return fibonacciNumbers;
}
//Math Solution: Extra space O(1) Runtime O(n)
List<Integer> getFibonacciNumbers(int[] nums) {
List<Integer> fibonacciNumbers = new ArrayList<>();
for(int n: nums) {
if(isFibonacci(n)) fibonacciNumbers.add(n);
}
return fibonacciNumbers;
}
boolean isFibonacci(int n)
{
// n is Fibonacci if one of 5*n*n + 4 or 5*n*n  4 or both
// is a perfect square
return isPerfectSquare(5*n*n + 4) 
isPerfectSquare(5*n*n  4);
}
boolean isPerfectSquare(int x)
{
int s = (int) Math.sqrt(x);
return (s*s == x);
}

aonecoding
February 15, 2018 Looking for coaching on interview preparation?
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SOLUTION:
final String[] largeNumbers = new String[] {""," thousand"," million"," billion"," trillion"," quintillion"};//...etc
final String[] digits = new String[] {"", " one", " two", " three", " four", " five", " six"," seven"," eight", " nine"};
final String[] tens = new String[] {"", " ten"," twenty"," thirty"," forty"," fifty"," sixty"," seventy"," eighty"," ninety"};
final String[] teens = new String[] {" ten", " eleven"," twelve", " thirteen", " fourteen", " fifteen", " sixteen", " seventeen"," eighteen"," nineteen"};
String digitsToEnglish(String num) {
if(num.length() == 1) {
return digits[Integer.parseInt(num)];
}
int len = num.length();
StringBuilder english = new StringBuilder();
int largeNumIdx = 0, tripletIdx = 0;
while(tripletIdx < len) {
StringBuilder triplet = new StringBuilder();
if(len > tripletIdx) {
triplet.append(num.charAt(len  1  tripletIdx));
}
if(len > tripletIdx + 1) {
triplet.insert(0, num.charAt(len  1  tripletIdx  1));
}
if(len > tripletIdx + 2) {
triplet.insert(0, num.charAt(len  1  tripletIdx  2));
}
if(Integer.parseInt(triplet.toString()) != 0) {
StringBuilder current = new StringBuilder();
if (triplet.length() == 3 && triplet.charAt(0) != '0') {
//current.append(' ');
current.append(digits[triplet.charAt(0)  '0']);
current.append(' ');
current.append("hundred");
}
if (triplet.length() >= 2 && triplet.charAt(triplet.length()  2) == '1') {
//current.append(' ');
current.append(teens[triplet.charAt(triplet.length()  1)  '0']);
} else {
if (triplet.length() >= 2) {
//current.append(' ');
current.append(tens[triplet.charAt(triplet.length()  2)  '0']);
}
//current.append(' ');
current.append(digits[triplet.charAt(triplet.length()  1)  '0']);
}
//current.append(' ');
current.append(largeNumbers[largeNumIdx]);
english.insert(0, current);
}
largeNumIdx += 1;
tripletIdx += 3;
}
return english.toString();
}

aonecoding
February 15, 2018 Looking for interview experience sharing and coaching?
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SOLUTION:
Twoway BFS, similar to 'WordLadder'.
s1Derives and s2Derives are the words reachable by swapping characters in s1 and s2 in any random way. If s1and s2 are anagrams, for sure s1Derives and s2Derives would join at some point. Find out all possible derivatives 'nextLevel' of s1/s2. Then find out all derivatives of words in 'nextLevel'......until the earliest joint of s1Derives and s2Derives occurs. Return the number of levels gone through.
int minNumberOfSwaps(String s1,String s2) {
//if(!isAnagram(s1, s2)) return Integer.MAX_VALUE; //assume s1 and s2 are anagrams
int step = 0;
Set<String> s1Derives = new HashSet<>();
Set<String> s2Derives = new HashSet<>();
s1Derives.add(s1);
s2Derives.add(s2);
Set<String> visited = new HashSet<>();
int len = s1.length();
while(!containsSameString(s1Derives, s2Derives)) {
Set<String> set = step % 2 == 0 ? s1Derives: s2Derives;
Set<String> nextLevel = new HashSet<>();
for(String s: set) {
for (int i = 0; i < len; i++) {
for (int j = i; j < len; j++) {
if(s.charAt(i) != s.charAt(j)) {
char[] charArray = swap(s.toCharArray(), i, j);
String derived = new String(charArray);
if(!visited.contains(new String(derived))) {
nextLevel.add(derived);
visited.add(derived);
}
}
}
}
}
if(step % 2 == 0) s1Derives = nextLevel;
else s2Derives = nextLevel;
step++;
}
return step;
}

aonecoding
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SOLUTION 1
public Integer random(int[] weights) {
int totalWeight = 0;
Integer selected = null;
Random rand = new Random();
for(int i = 0; i < weights.length; i++) {
int r = rand.nextInt(totalWeight + weights[i]);
if(r >= totalWeight) {
selected = i;
totalWeight += weights[i];
}
}
return selected;
}
FOLLOWUP
//main
//RandomSelectTree tree = new RandomSelectTree();
//while(tree.n > 0) tree.randomPoll();
//Segment Tree O(nlogn)
public class RandomSelectTree {
int[] tree;
int[] weights;
int n; //number of remaining balls
RandomSelectTree(int[] weights) {
tree = new int[2 * weights.length + 1];
this.weights = weights;
n = weights.length;
buildTree(0, 0, weights.length  1);
}
//build segment tree for weights[]
private void buildTree(int root, int start, int end) {
if(start > end) return;
if(start == end) {
tree[root] = weights[start];
return;
}
int mid = (start + end) / 2;
int leftChild = 2 * root + 1, rightChild = 2 * root + 2;
buildTree(leftChild, start, mid);
buildTree(rightChild, mid + 1, end);
tree[root] = tree[leftChild] + tree[rightChild];
}
public int randomPoll() {
if(n == 0) {
//Exception
}
int rand = new Random().nextInt(tree[0]);
int start = 0, end = weights.length  1, node = 0;
while(start < end) {
//chosen ball is between start to (start + end) / 2
if(rand < tree[node * 2 + 1]) {
node = node * 2 + 1;
end = (start + end) / 2;
} else { //chosen ball between (start + end) / 2 + 1 and end
rand = tree[node * 2 + 1]; //!!! narrow down to remaining range
node = node * 2 + 2;
start = (start + end) / 2 + 1;
}
}
delete(start, node);
return start;
}
//remove ball at idx.
private void delete(int idx, int node) {
n;
while(node > 0) {
tree[node] = weights[idx];
node = (node  1) / 2;
}
tree[0] = weights[idx];
}

aonecoding
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//Solution to question O(n)
boolean isPeriod(String s) {
StringBuilder str = new StringBuilder(s + s);
str.deleteCharAt(0);
str.deleteCharAt(str.length()  1);
return strStr(str.toString(), s); //KMP strStr(T, S) to find if T has S in it.
}
//Solution to followup
//This method looks for the repeating pattern in string
private static String getPeriod(String string) { // O(n * n)
//for every possible period size i, check if it's valid
for (int i = 1; i <= string.length() / 2; i++) {
if (string.length() % i == 0) {
String period = string.substring(0, i);
int j = i;
while(j + i < string.length()) {
if (period.equals(string.substring(j, j + i))) {
j = j + i;
if(j == string.length()) { //period valid through entire string
return period;
}
} else {
break;
}
}
}
}
return null; //string is not periodic
}

aonecoding
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SOLUTION:
Step 1, have a table to store all the visits sorted by time stamp.
or
have a queue to store the visits per second in the past 5 minute.
Followup:
Have two arrays hits[range] and lastupdated[range].
Range = 5 mins = 300 seconds in this case. Any hit within 'range' time from now is valid and should counted.
Index of a hit in the two arrays will be timestamp % 300.
Store the last updated time of a hit. Later if a query for the hit count arrives with a query timestamp, sum up all the valid hits from array 'hits'. Threshold for valid hits: query time  hit last updated time < 300.
public class HitCounter {
int[] times, hits;
int timeRange;
/** Initialize your data structure here. */
public HitCounter(int range) {
times = new int[range];
hits = new int[range];
timeRange = range;
}
/** Record a hit.
@param timestamp  The current timestamp (in seconds granularity). */
public void hit(int timestamp) {
int idx = timestamp % timeRange;
if (times[idx] != timestamp) {
times[idx] = timestamp;
hits[idx] = 1;
} else {
++hits[idx];
}
}
/** Return the number of hits in the past 5 minutes.
@param timestamp  The current timestamp (in seconds granularity). */
public int getHits(int timestamp) {
int res = 0;
for (int i = 0; i < timeRange; ++i) {
if (timestamp  times[i] < timeRange) {
res += hits[i];
}
}
return res;
}
}
Followup2
For writing,
Concurrency update becomes an issue. Add write lock for protection. But this may slowdown the machine badly.
Move hit counters onto a distributed system. Have several machines counting together. Assign userIDs to diff hosts.
Add LB on top to make sure requests get distributed evenly.
Upon reading, sum up counts across all machines. For a readheavy system, add cache.
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//Binary Indexed Tree (O(logN) Toggle, O(logN) Get.
public class ToggleBulbs {
int[] bulbs;
public ToggleBulbs(int n) { //n bulbs given
bulbs = new int[n + 1];
}
public boolean isOn(int i) {
//a bulb is on if it's toggled an odd number of times
return read(i) % 2 == 1;
}
// toggle(i, j) is equivalent to
// toggle(i, n  1) and then toggle(j, n  1)
public void toggle(int i,int j) {
toggle(i);
toggle(j + 1);
}
//toggle from ith to the last bulb (a standard update in BITree)
private void toggle(int idx){
int node = idx + 1;
while (node < bulbs.length){
bulbs[node] = (bulbs[node] + 1) % 2;
node += (node & node);
}
}
//get the number of times bulb i toggled (read prefix sum from 0 to i in BITree)
private int read(int idx) {
int node = idx + 1;
int sum = 0;
while (node > 0) {
sum += bulbs[node];
node = (node & node);
}
return sum;
}
}

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boolean hasFactorKSubArray(int[] arr, int k) {
for(int i = 1; i < arr.length; i++) {
arr[i] += arr[i  1]; //becomes a prefix sum array of arr (can be recovered if necessary)
}
for(int hi = arr.length  1; hi >= 1; hi ) {
for(int lo = 0; lo <= hi  1; lo++) {
if((arr[hi]  lo == 0 ? 0: arr[lo  1]) % k == 0) {
return true;
}
}
}
return false;
}

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class Node:
def __init__(self, ID, parent):
self.ID = ID
self.parentID = parent
self.left = None
self.right = None
#function to identify if given is preorder
'''
Create a stack to store nodes in the current path when traversing.
Push node[i] into stack once node[i] is verified to be valid (valid only when parent of node[i] is in stack.
In preorder a parent must show up earlier than its child)
Whenever stack top is not the parent of node[i], pop until parent of node[i] is at stack top. Push node[i].
If all nodes popped but parent of node[i] still not found, then node[i] is not in preorder sequence.
'''
def isPreorder(nodes):
if not nodes:
return True
st = [nodes[0].ID]
i = 1
while i < len(nodes):
if not st:
return False
if st[1] is nodes[i].parentID:
st.append(nodes[i].ID)
i += 1
else:
st.pop()
return True

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# The question is about identifying any window size k that its numbers at both ends bigger than numbers in between
def bloomDay(a, k): # k must be >= 0
window = [] # elements in window maintain descending order.
firstDay = 10000000
i = 0
while i < len(a):
if not window:
window.append(i)
i += 1
elif i  window[0] <= k:
if a[i] >= a[window[0]]:
window = [i]
i += 1
elif a[i] >= a[window[1]]:
window.pop()
else:
window.append(i)
i += 1
else: #window size == k
if len(window) == 1 or a[window[1]] < a[i]:#valid window with both end numbers biggest
firstDay = min(firstDay, a[i], a[window[0]])
window = [i]
i += 1
else: #a[i] is less than a number in the middle of the window
window.pop(0)
if a[window[1]] <= a[i]:
window.pop()
else:
window.append(i)
i += 1
return firstDay

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SOLUTION
The push, pop and inc can all take place in O(1) time
if there is an additional array to maintain the m incremented at each position n.
class Stack:
def __init__(self):
self.nums = []
self.add = []
def push(self,num):
self.nums.append(num)
self.add.append(0)
print num," "
def pop(self):
try:
number_to_add = self.add.pop()
print self.nums.pop() + number_to_add
if self.add:
self.add[1] += number_to_add
except:
print "can't pop from an empty stack"
def inc(self, n, m):
if not self.nums:
print "empty"
if n > 0:
n = min(n, len(self.nums))
self.add[n  1] += m
print self.nums[1] + self.add[1], " "

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Sliding window anagram:
public static List<Integer> getAnagrams(String s, String word) {
Map<Character, Integer> letters = new HashMap<>();
int distinct_letters = 0;
for(char c: word.toCharArray()) {
if(!letters.containsKey(c)) distinct_letters++;
letters.put(c, letters.getOrDefault(c, 0) + 1);
}
//search for anagrams with two pointers
List<Integer> res = new ArrayList<>();
int lo = 0, hi = 0;
while(hi < s.length()) {
if(!letters.containsKey(s.charAt(hi))) {
while(lo < hi) {
char c = s.charAt(lo);
if(letters.get(c) == 0) distinct_letters++;
letters.put(c, letters.get(c) + 1);
lo++;
}
hi++;
lo = hi;
} else if(letters.get(s.charAt(hi)) == 0){
while(s.charAt(lo) != s.charAt(hi)) {
char c = s.charAt(lo);
if(letters.get(c) == 0) distinct_letters++;
letters.put(c, letters.get(c) + 1);
lo++;
}
lo++;
} else {
char c = s.charAt(hi);
letters.put(c, letters.get(c)  1);
if(letters.get(c) == 0) distinct_letters;
if(distinct_letters == 0) {
res.add(lo);
}
hi++;
}
}
return res;
}

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Get sqrt(x):
int sqrt(int x) {
//check for x >= 0 if necessary
int r = x;
while (r * r > x)
r = (r + x / r) / 2;
return r ;
}
BST Implementation:
class BinarySearchTree {
/* Class containing left and right child of current node and key value*/
class Node {
int key;
Node left, right;
public Node(int item) {
key = item;
left = right = null;
}
}
// Root of BST
Node root;
// Constructor
BinarySearchTree() {
root = null;
}
// This method mainly calls insertRec()
void insert(int key) {
root = insertRec(root, key);
}
/* A recursive function to insert a new key in BST */
Node insertRec(Node root, int key) {
/* If the tree is empty, return a new node */
if (root == null) {
root = new Node(key);
return root;
}
/* Otherwise, recur down the tree */
if (key < root.key)
root.left = insertRec(root.left, key);
else if (key > root.key)
root.right = insertRec(root.right, key);
/* return the (unchanged) node pointer */
return root;
}
// This method mainly calls InorderRec()
void inorder() {
inorderRec(root);
}
// A utility function to do inorder traversal of BST
void inorderRec(Node root) {
if (root != null) {
inorderRec(root.left);
System.out.println(root.key);
inorderRec(root.right);
}
}
// This method mainly calls deleteRec()
void deleteKey(int key)
{
root = deleteRec(root, key);
}
/* A recursive function to insert a new key in BST */
Node deleteRec(Node root, int key)
{
/* Base Case: If the tree is empty */
if (root == null) return root;
/* Otherwise, recur down the tree */
if (key < root.key)
root.left = deleteRec(root.left, key);
else if (key > root.key)
root.right = deleteRec(root.right, key);
// if key is same as root's key, then This is the node
// to be deleted
else
{
// node with only one child or no child
if (root.left == null)
return root.right;
else if (root.right == null)
return root.left;
// node with two children: Get the inorder successor (smallest
// in the right subtree)
root.key = minValue(root.right);
// Delete the inorder successor
root.right = deleteRec(root.right, root.key);
}
return root;
}
}

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Get the longest arithmetic progression:
public static int[] llac(int[] array) {
if(array.length == 0) return new int[]{};
int maxLen = 1, //len of longest sequence
end = array[0], //end number of sequence
step = 0;
Map<Integer,Map<Integer, Integer>> index_to_llac = new HashMap<>(); //Map<currentIndex, Map<difference, length>>
for(int current = 1; current < array.length; current++) {
index_to_llac.put(current, new HashMap<>());
for(int prev = 0; prev < current; prev++) {
int diff = array[current]  array[prev];
int length = 2;
if(index_to_llac.containsKey(prev) && index_to_llac.get(prev).containsKey(diff)) {
length = index_to_llac.get(prev).get(diff) + 1;
}
Map<Integer, Integer> diff_to_llac = index_to_llac.get(current);
//if this is a arithmetic progression with [diff] exists before current, add 1 to the length to include current in the progression
//otherwise, [prev, current] forms a progression of size 2
diff_to_llac.put(diff, length);
if(maxLen < length) {
maxLen = length;
end = array[current];
step = diff;
}
}
}
int[] sequence = new int[maxLen];
for(int i = maxLen  1; i >= 0; i) {
sequence[i] = end;
end = step;
}
return sequence;
}

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public static int rankWays(int n) {
int[] base = new int[2]; // start from case for 0 participants
base[1] = 1;
for(int p = 2; p <= n; p++) {
int[] ways = new int[p + 1]; //index is row number, value is count of ways
for(int r = 1; r < base.length; r++) {
ways[r + 1] += base[r] * (r + 1);
ways[r] += base[r] * r;
}
base = ways;
}
int sum = 0;
for(int x: base) sum += x;
return sum;
}
With N participants [p1, p2, p3... pN], one ranking could be:
1st place: p1
2nd place: p2, p3
...
xth place: pN
In this case, if one more player pL joins the game. The new player pL can be placed into
pL can be here as new 1st place
1st place p1 pL can be here tied with 1st place
pL can be here as new 2nd place
2nd place p2, p3 pL can be here tied 2nd place
... ...
xth place pN pL can be here tied last place
pL can be here as new last
So for every rank combination of N people, adding 1 more player creates rowNumber * 2 + 1 ways of ranking.
So if we keep track of the row numbers of all previous combinations of ranking,
a ranking with R rows is gonna derive into R + 1 more cases (with R + 1 row numbers) plus R more cases (with R row numbers).
To print all possible combinations:
public static List<List<List<Integer>>> rankCombinations(int n) {
List<List<List<Integer>>> base = new ArrayList<>();
base.add(new ArrayList<>());
base.get(0).add(new ArrayList<>());
base.get(0).get(0).add(1);
for(int p = 2; p <= n; p++) {
List<List<List<Integer>>> combinations = new ArrayList<>();
for(List<List<Integer>> comb: base) {
for(int i = 0; i <= comb.size(); i++) {
List<Integer> rank = new ArrayList<>();
rank.add(p);
List<List<Integer>> newComb = new ArrayList<>(comb);
newComb.add(i, rank);
combinations.add(newComb);
}
for(int i = 0; i < comb.size(); i++) {
List<List<Integer>> newComb = deepcopy(comb);
newComb.get(i).add(p);
combinations.add(newComb);
}
}
base = combinations;
}
//print all ranking combinations. multi players in a same sublist makes a tie.
for(List<List<Integer>> ranks: base) System.out.println(ranks);
return base;
}
private static List<List<Integer>> deepcopy(List<List<Integer>> ls) {
List<List<Integer>> copy = new ArrayList<>();
for(List<Integer> sublist: ls) {
copy.add(new ArrayList<Integer>(sublist));
}
return copy;
}
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Candidate’s Solution (Passed all test cases in Airbnb OA)
int paginate(const std::vector<std::string>& addressbook, std::vector<std::string>& results) {
std::list<std::pair<int, int> > curve; // host_id, position in addressbook
int num_entries = addressbook.size(); // already sorted by score.
for (int index = 0; index < num_entries; ++index) {
const std::string& s = addressbook[index];
int host_id = std::stoi(s.substr(0, s.find_first_of(',')));
curve.emplace_back(host_id, index);
}
while (!curve.empty()) {
std::unordered_set<int> uniq_host_ids;
std::vector<decltype(curve)::iterator> hosts;
bool combo = false;
for (auto iter = curve.begin(); iter != curve.end(); ++iter) {
if (uniq_host_ids.find(iter>first) == uniq_host_ids.end()) {
uniq_host_ids.insert(iter>first);
hosts.push_back(iter);
if (hosts.size() == resultsPerPage) {
combo = true;
break;
}
}
}
for (auto& host : hosts) {
results.push_back(addressbook[host>second]);
}
for (auto& host : hosts) {
curve.erase(host);
}
if (combo) {
if (!curve.empty()) results.push_back("");
continue; // enter next loop
}
if (!curve.empty()) {
int num_done = hosts.size();
for (auto iter = curve.begin(); iter != curve.end(); ++iter) {
hosts.push_back(iter);
if (hosts.size() == resultsPerPage) {
combo = true; break;
}
}
int num_hosts = hosts.size();
for (int i = num_done; i < num_hosts; ++i) {
results.push_back(addressbook[hosts[i]>second]);
}
for (int i = num_done; i < num_hosts; ++i) {
curve.erase(hosts[i]);
}
if (combo && !curve.empty()) {
results.push_back("");
}
} // end if condition
} // end while loop
return 0;
}

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public static List<Integer> getAnagrams(String s, String word) {
Map<Character, Integer> letters = new HashMap<>();
int distinct_letters = 0;
for(char c: word.toCharArray()) {
if(!letters.containsKey(c)) distinct_letters++;
letters.put(c, letters.getOrDefault(c, 0) + 1);
}
//search for anagrams with two pointers
List<Integer> res = new ArrayList<>();
int lo = 0, hi = 0;
while(hi < s.length()) {
if(!letters.containsKey(s.charAt(hi))) {
while(lo < hi) {
char c = s.charAt(lo);
if(letters.get(c) == 0) distinct_letters++;
letters.put(c, letters.get(c) + 1);
lo++;
}
hi++;
lo = hi;
} else if(letters.get(s.charAt(hi)) == 0){
while(s.charAt(lo) != s.charAt(hi)) {
char c = s.charAt(lo);
if(letters.get(c) == 0) distinct_letters++;
letters.put(c, letters.get(c) + 1);
lo++;
}
lo++;
} else {
char c = s.charAt(hi);
letters.put(c, letters.get(c)  1);
if(letters.get(c) == 0) distinct_letters;
if(distinct_letters == 0) {
res.add(lo);
}
hi++;
}
}
return res;
}

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public static void main(String[] args) {
System.out.println(lonelyPixelCount(new int[][]{ //1: black, 0: white
{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0},
{0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0},
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0},
{0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
}));
}
static int lonelyPixelCount(int[][] picture) {
int m = picture.length, n = picture[0].length;
//First traversal sum up the count of black pixels by column
for(int i = 1; i < m; i++){
for(int j = 0; j < n; j++){
picture[i][j] += picture[i  1][j];
}
}
int result = 0;
//Then traverse row by row from the bottom, count the black pixels in current row.
//If there is only 1 black pixel in current row, verify whether it is also the only in the column.
for(int i = m  1; i >= 0; i) {
int pixel_count_in_row = 0;
boolean only_pixel_in_column = false;
for(int j = n  1; j >= 0; j) {
if(picture[i][j] > 0 && (i == 0  picture[i  1][j] + 1 == picture[i][j])) { //verify if current cell number is a black pixel, by method in blue text above
pixel_count_in_row++;
if((i < m  1 && picture[i + 1][j] == 1)  (i == m  1 && picture[i][j] == 1)) {
only_pixel_in_column = true;
}
}
if(i < m  1) {
//overwrite current cell with the number below it
picture[i][j] = picture[i + 1][j];
}
}
if(pixel_count_in_row == 1 && only_pixel_in_column) {
result++;
}
}
return result;
}

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import java.util.List;
import java.util.ArrayList;
public class IntegerCollection {
List<Integer> collection = new ArrayList<>();
List<Integer> divisors = new ArrayList<>();
int factor = 1;
int addition = 0;
int set0 = 1;
public void append(int x) {
collection.add(x  addition);
divisors.add(factor);
}
public int get(int index) {
if(index >= collection.size()) throw new RuntimeException();
if(divisors.get(index) <= set0) return addition;
return collection.get(index) * (factor / divisors.get(index)) + addition;
}
public void add_to_all(int x) {
addition += x;
}
public void multiply_to_all(int x) {
if(x == 0) {
addition = 0;
factor = 1;
set0 = collection.size()  1;
} else {
addition *= x;
factor *= x;
}
}
}

aonecoding
September 22, 2017 SELECT
Department.name,
COUNT(Employee.id)
FROM
Department
LEFT JOIN
Employee ON Department.dept_id = Employee.dept_id
GROUP BY
Department.dept_id,
Department.name
ORDER BY
COUNT(Employee.id) DESC,
Department.name
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Looking for interview experience sharing and coaching?
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Our ONE TO ONE class offers
SYSTEM DESIGN Courses (highly recommended for candidates for FLAG & U)
ALGORITHMS (conquer DP, Greedy, Graph, Advanced Algos & Clean Coding),
latest interview questions sorted by companies,
mock interviews.
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Solution to III.
int sqrt(int x) {
if (x == 0)
return x;
int left = 1, right = x;
while (true) {
int mid = (left + right) / 2;
if (mid > x / mid)
right = mid  1;
else if (mid + 1 > x / (mid + 1)) //mid < x / mid
return mid;
else //mid < x / mid
left = mid + 1;
}
}
IV. is on LC
V. Lossy Counting & Sticky Sampling
Solution to Closest K Nodes
vector<int> closestKValues(TreeNode* root, double target, int k) {
vector<int> res;
inorder(root, target, k, res);
return res;
}
void inorder(TreeNode *root, double target, int k, vector<int> &res) {
if (!root) return;
inorder(root>left, target, k, res);
if (res.size() < k) res.push_back(root>val);
else if (abs(root>val  target) < abs(res[0]  target)) {
res.erase(res.begin());
res.push_back(root>val);
} else return;
inorder(root>right, target, k, res);
}
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Solution:
How to combine the 4 given card ranks with operators +, , x and / to reach 24.
Backtrack to print all possible combinations.
public void solve(int[] cards) {
boolean[] played = new boolean[4];
for(int i = 0; i < 4; i++) {
StringBuilder str = new StringBuilder(String.valueOf(cards[i]));
played[i] = true;
solve(cards, played, cards[i],str);
played[i] = false;
}
}
private void solve(int[] cards, boolean[] played, double res, StringBuilder str) {
if(played[0] && played[1] && played[2] && played[3]) {
if(res == 24)
System.out.println(str.toString());
return;
}
for(int i = 0; i < 4; i++) {
int len = str.length();
if(!played[i]) {
played[i] = true;
str.insert(0, '(');
str.append("+" + cards[i] + ")");
solve(cards, played, res + cards[i], str);
str.setCharAt(len + 1, '');
solve(cards, played, res  cards[i], str);
str.deleteCharAt(0);
str.deleteCharAt(str.length()  1);
str.setCharAt(len, '*');
solve(cards, played, res * cards[i], str);
str.setCharAt(len, '/');
solve(cards, played, 1.0 * res / cards[i], str);
str.delete(len, str.length());
played[i] = false;
}
}
}

aonecoding
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Solution:
I. If it is a writeheavy application, create a Fenwick Tree to store for every toggle(i, j), add 1 to node i and 1 to node j (O(logN)). Upon calling isOn(k), get the cumulative sum (O(logN)) till the kth node. If sum is odd, then bulb is on. Otherwise bulb is off.
II. If it is readheavy, create a bit map that toggles in linear time and reads in constant time.
//Fenwick Tree Solution
public class ToggleBulbs {
int[] sum;
public ToggleBulbs(int n) {
sum = new int[n];
}
public boolean isOn(int i)
{
return read(i)%2==1;
}
public void toggle(int start,int end)
{
update(start,1);
update(end+1,1);
}
private int read(int idx){
int ans = 0;
while (idx > 0){
ans += sum[idx];
idx = (idx & idx);
}
return ans;
}
private void update(int idx ,int val){
while (idx <= sum.length){
sum[idx] += val;
idx += (idx & idx);
}
}
}

aonecoding
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public void powerOf2Paths(int N) {
List<Integer> path = new ArrayList<>();
path.add(0);
powerOf2PathsHelper(4, 0, path);
}
public void powerOf2PathsHelper(int N, int num, List<Integer> path) {
if(num == N) {
System.out.println(path);
}
for(int i = 0; num + (1 << i) <= N; i++) {
int sum = num + (1 << i);
path.add(sum);
powerOf2PathsHelper(N, sum, path);
path.remove(path.size()  1);
}
}

aonecoding
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//get the number of connected components in a list
public int numberOfGroups(Set<Node> nodes) {
Set<Node> visited = new HashSet<>();
int cnt = 0;
for(Node node: nodes) {
while(!visited.contains(node)) {
if(nodes.contains(node)){
visited.add(node);
node = node.next;
} else {
cnt++;
break;
}
}
}
return cnt;
}

aonecoding
August 09, 2017 Hi Fernando,
I interpret the problem this way  making someone my colleague means having him into my team (assigning my manager to him). Since it is a design problem it's openended and your solution would be good as well.
Thanks for the reply!
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Solution:
Breath First Search.
Create a map that stores Map<Node, [PreviousNode, Cost]> the cost to the current node.
Since the question asks for the actual shortest path, in the map also store the previous node in the path.
During the BFS, if the current node has been visited before, but the current cost < former cost, update the cost and route for the current node (only keep the min cost route). Otherwise if current cost > former cost, stop search by not including the current node into the BFS queue.
To stand for a node in the map with an integer rather than a point (x, y), do
nodeID = x * matrix[0].length + y;
To convert the ID back to a point, do
x = nodeID / matrix[0].length;
y = nodeID % matrix[0].length;
//Followup
public void minCostPath(int[][] matrix) {
Queue<Integer> queue = new ArrayDeque<>();
Map<Integer, int[]> map = new HashMap<>(); //key: current location; value[0]: the previous node from where it gets to the current location, value[1]:total cost up to current node
for(int j = 0; j < matrix[0].length; j++) { //put first row into queue
if(matrix[0][j] != 0) {
queue.add(j);
map.put(j, new int[] {1, matrix[0][j]});
}
}
int destination = 1, minCost = Integer.MAX_VALUE;
while(!queue.isEmpty()) {
int id = queue.poll();
int fromX = id / matrix[0].length, fromY = id % matrix[0].length;
if(fromX + 1 < matrix.length) {
int cost = moveMinCost(queue, map, matrix, id, fromX + 1, fromY);
if (cost >= 0 && fromX + 1 == matrix.length  1) {
if (cost < minCost) {
destination = id + matrix[0].length;
minCost = cost;
}
}
}
if(fromY + 1 < matrix[0].length)
moveMinCost(queue, map, matrix, id, fromX, fromY + 1);
if(fromX  1 >= 0)
moveMinCost(queue, map, matrix, id, fromX  1, fromY);
if(fromY  1 >= 0)
moveMinCost(queue, map, matrix, id, fromX, fromY  1);
}
if(destination == 1) return; //no such path from first row to last row
while(map.containsKey(destination)) { //print shortest path from destination to source
int x = destination / matrix[0].length, y = destination % matrix[0].length;
System.out.println("(" + x + ","+ y + "),");
destination = map.get(destination)[0];
}
}
private int moveMinCost(Queue<Integer> queue, Map<Integer, int[]> map, int[][] matrix, int from, int x, int y) {
if(matrix[x][y] == 0) return 1;
int id = x * matrix[0].length + y;
int cost = map.get(from)[1] + matrix[x][y];
if(!map.containsKey(id)  map.get(id)[1] > cost) {
map.put(id, new int[]{from, cost});
queue.add(id);
return cost;
}
return 1;
}

aonecoding
August 03, 2017 class Employee {
int id;
private String name;
//...other personal information
private Employee manager;
private List<Employee> subordinates; //direct subordinates
public Employee(int id, String name) {
this.id = id;
this.name = name;
subordinates = new ArrayList<>();
}
boolean isManager(Employee manager) {
Employee upperLevel = this.manager;
while(upperLevel != null && upperLevel != manager) {
upperLevel = upperLevel.manager;
}
return upperLevel.manager == manager;
}
void beColleague(Employee p) {
p.setManager(this.manager);
}
void setManager(Employee m) {
if(manager != null) { //remove from subordinate's list of current manager
manager.deleteSubordinate(this);
}
manager = m;
m.addSubordinate(this); //add to new manager's subordinate's list
}
private void deleteSubordinate(Employee m) {
subordinates.remove(m);
}
private void addSubordinate(Employee m) {
subordinates.add(m);
}
}
public class Employment {
private Employee admin;
private Map<Integer, Employee> employees; //id: employee
public Employment() {
admin = new Employee(0, "ADMINISTRATOR"); //root of the employment tree, the highest level supervisor
employees = new HashMap<>();
employees.put(0, admin);
}
public void assignManager(int p1, int p2) {
employees.get(p2).setManager(employees.get(p1));
}
public void beColleague(int p1, int p2) {
employees.get(p2).beColleague(employees.get(p1));
}
public boolean isManager(int p1, int p2) {
return employees.get(p2).isManager(employees.get(p1));
}
//public void addEmployee(int p);
//public void deleteEmployee(int p);
}
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Solution to followup: Randomly select a rectangle based on the areas as the weights (larger weights leads to higher probability). Then randomly find a point from the chosen rectangle.
public class RandomSelectFromRectangle {
class Rectangle {
int x1, x2, y1, y2; //top left (x1, y1), bottom right (x2, y2)
}
class Point {
int x, y;
public Point(int a, int b) {
x = a;
y = b;
}
}
final Random rand = new Random();
//Round2
public Point randomSelectFrom(Rectangle rectangle) {
return new Point(rectangle.x1 + rand.nextInt(rectangle.x2  rectangle.x1 + 1),
rectangle.y2 + rand.nextInt(rectangle.y1  rectangle.y2 + 1));
}
//Round2 followup
//first of all decide which rectangle yields the point (randomly select a rectangle based on area as the weight)
//then apply randomSelectFrom(rectangle) on the selected rectangle
public Point randomSelectFrom(Rectangle[] rectangles) {
int selected = 1, total = 0;
for(int i = 0; i < rectangles.length; i++) {
int area = (rectangles[i].x2  rectangles[i].x1) * (rectangles[i].y1  rectangles[i].y2);
if(rand.nextInt(total + area) >= total) {
selected = i;
}
total += area;
}
return randomSelectFrom(rectangles[selected]);
}
}

aonecoding
August 03, 2017 It does not matter. The start time and the end time of intervals are in equal position for this problem.
e.g.
[1, 5], [2, 10], [6, 9] is currently sorted by start. You may merge them from left to right and get the result [1,10].
If they were sorted by end time instead  [1, 5], [6, 9],[2, 10]. You could merge them from right to left and get the same result [1, 10].
It just depends on if you wanna start the merge from the left or right side of the given inteval list.
A sample code for when intervals sorted by start time
public List<Interval> merge(List<Interval> intervals) {
if (intervals.size() <= 1)
return intervals;
// Sort by ascending starting point
intervals.sort((i1, i2) > Integer.compare(i1.start, i2.start));
List<Interval> result = new LinkedList<Interval>();
int start = intervals.get(0).start;
int end = intervals.get(0).end;
for (Interval interval : intervals) {
if (interval.start <= end) // Overlapping intervals, move the end if needed
end = Math.max(end, interval.end);
else { // Disjoint intervals, add the previous one and reset bounds
result.add(new Interval(start, end));
start = interval.start;
end = interval.end;
}
}
// Add the last interval
result.add(new Interval(start, end));
return result;
}
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Solution:
 aonecoding August 24, 2018