CareerCup

In 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.
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



Follow-up:
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.

In 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.
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



Follow-up:
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.

Find three non-overlap windows of size k in an int array, that together has a maximum sum
of the 3k entries.
example
int[] nums = [1,2,1,2,6,7,5,1]
k = 2
output
[1,2],[2,6],[7,5]

Given:
a encoded to 1
b encoded to 2
....
z encoded to 26

You can translate a number to a string:

'123' can be translated to 'abc'
but also can be translated to 'aw','lc' which gives 3 total translations
'12' can be translated to 'ab' and 'l' -> 2 translations

Write a function to get the number of valid combinations from a number like '123123123'

how to find leaf node value from preorder sequence of BST without rebuilding the tree

deleted

for a binary tree, print the root to the leaf path, but add "_" to indicate the relative position.
example:

A 
       /   \  
      B     C 
     / \   / \
    D  E   F  G

output
_ _ A
_ B
D

_A
B
_E

A
_C
F

A
_ C
_ _ G

Q: 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 non-overlap meetings.

class Meeting:
def __init__(self):
	self.startTime
	self.endTime
	self.weight

@concernedCoder
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?

If a string is matched to any filter, it is in the black list, otherwise not.
Design a data structure and implement following two functions.

addFilter(filter)
isInBlackList(string)

filters are in the form of
“a*b”
“abc”
“aa*b”
having at most one star, which matches 0 or more chars.

You have L, a list containing some digits (0 to 9). Write a function answer(L) which finds the largest number that can be made from some or all of these digits and is divisible by 3. If it is not possible to make such a number, return 0 as the answer. L will contain anywhere from 1 to 9 digits. The same digit may appear multiple times in the list, but each element in the list may only be used once.

{{
Test cases
==========

Inputs:
(int list) l = [3, 1, 4, 1]
Output:
(int) 4311

Inputs:
(int list) l = [3, 1, 4, 1, 5, 9]
Output:
(int) 94311
}}

My Solution:
{{

package com.google.challenges;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;


public class Answer {
public static int answer(int[] l) {

// Your code goes here.
ArrayList<Integer> list0 = new ArrayList<>();
ArrayList<Integer> list1 = new ArrayList<>();
ArrayList<Integer> list2 = new ArrayList<>();
int sum =0;
Arrays.sort(l);
for(int i = 0; i<l.length; i++){
if(l[i] % 3 == 0){
list0.add(l[i]);
}else if(l[i] % 3 == 1){
list1.add(l[i]);
}else{
list2.add(l[i]);
}
sum += l[i];
}

if(sum%3==0){
StringBuilder strNum = new StringBuilder();
for(int i = l.length-1; i >= 0; i--)
{
strNum.append(l[i]);
}
return Integer.parseInt(strNum.toString());
}else if(sum%3 == 1){
if(list1.size()>0){
Collections.sort(list1);
list1.remove(0);
}else if(list2.size() >= 2){
Collections.sort(list2);
list2.remove(1);
list2.remove(0);
}else{
return -1;
}
}else if(sum%3 == 2){
if(list2.size()>0){
Collections.sort(list2);
list2.remove(0);
}else if(list1.size() >= 2){
Collections.sort(list1);
list1.remove(1);
list1.remove(0);

}else{
return -1;
}
}
list0.addAll(list1);
list0.addAll(list2);
StringBuilder strNum = new StringBuilder();
Collections.sort(list0);
for(int i = list0.size()-1; i >= 0; i--)
{
strNum.append(list0.get(i));
}
return strNum.length() > 0 ? Integer.parseInt(strNum.toString()) : -1;
}
}
}}

But here I am able to pass 4 test cases out of 5. Therefore I am looking for scenario which is left to check.
Can someone help me?

On google search, how to enable key word auto completion after a few letters typed.

Follow-up: How to rank the words if they are weighted by frequency?

Cross the street
ABC Company is involved in the logistics business.

The company has many outlets and stockyards in a city. The city is like an
N
×
M
N×M grid. We consider a single cell of the given grid to be a single block in the city. The stockyard is at the upper-left corner and the outlet is located in the lower right corner. Everyday, one of the employees has to travel from the upper left to the lower right corner for supplies. Each block in the city has a height, where the height of the block located at position (i,j) in the grid is equal to
A
[
i
]
[
j
]
A[i][j]. The company wants to change the heights of some of the blocks, so that the employee can enjoy a high-speed drive from the stockyard to the outlet. But this change comes at a certain cost.

Specifically, if they change a block height from x to y, then they must pay exactly
|
x
−
y
|
|x−y| dollars. Please help them find the minimum cost, such that by spending that specific amount, they can get a path from stockyard to the outlet with all blocks along the path having the same height.

In a single move, the employee can move from a block to any of its adjacent blocks. Note that during this journey, the employee is allowed to move in all four directions, fulfilling the condition that he never goes out of the grid at any point in time.

Input :

First line contains two positive integers N and M - number of rows and columns in the city. Then, N lines follow, each containing M integers, where the
j
t
h
jth integer on the
i
t
h
ith line denotes
A
[
i
]
[
j
]
A[i][j].

Output :

The first and only line of output should contain minimum cost.

Constraints :

1<= N, M <=100

1<= height of blocks <=100

Sample Input
5 5
1 1 1 1 1
9 9 9 9 1
1 3 3 3 1
1 9 9 9 9
1 1 1 1 1
Sample Output
6
Explanation
Optimal path taken by the employee will be : (1,1) -> (1,2) -> (1,3) -> (1,4) -> (1,5) -> (2,5) -> (3,5) -> (3,4) -> (3,3) -> (3,2) -> (3,1) -> (4,1) -> (5,1) -> (5,2) -> (5,3) -> (5,4) -> (5,5) The height of each block along this path can be changed to
1
1, at a total cost of
6
6. There is no way to get a cost less than this.

Alex has recently decided to learn about how to design compilers. As a first step he needs to find the number of different variables that are present in the given code.


So Alex will be provided N statements each of which will be terminated by a semicolon(;). Now Alex needs to find the number of different variable names that are being present in the given statement. Any string which is present before the assignment operator denotes to a variable name.

Input Format: :

The first line contains a single integer N


Each of the next N lines contains a single statement.

It is guaranteed that all the given statements shall be provided in a valid manner according to the format specified.

Output Format: :

Print the number of different variable name that are present in the given statements.


Sample Input
2
foo = 3;
bar = 4;
Sample Output
2
Explanation
Foo and Bar are only two variables used inside the statements so answer is 2.

Design classes to represent the following problem and solve the questions 1,2,3
A user might have some outstanding auto loan amount and you have 3 types of offers: personal loan, credit card and auto loan offers. You need to provide the user
with the following details:
1. Send user all the offers to the user
2. Send user all eligible offers (where minCreditScore < userCreditScore < maxCreditScore)
3. Send user all offers which satisfied 2) and where the (userOutStandingLoanAmount < maxOfferedAutoLoanAmount)



personal-loan = [{
"personal-loan": {
"id": 1,
"provider": "Avant",
"term": 36,
"minimumCreditScore": 300,
"maximumCreditScore": 700,
"maximumAmount": 10000
}
}, {
"personal-loan": {
"id": 2,
"provider": "Prosper",
"term": 24,
"minimumCreditScore": 600,
"maximumCreditScore": 700,
"maximumAmount": 5000
}
}]

credit-card=[{
"credit-card": {
"id": 2,
"provider": "CapitalOne",
"minimumCreditScore": 600,
"maximumCreditScore": 700
}
}, {
"credit-card": {
"id": 3,
"provider": "Chase",
"minimumCreditScore": 300,
"maximumCreditScore": 900
}
}]

autoloan = [{
"auto-loan": {
"id": 1,
"provider": "CapitalOne",
"term": 36,
"minimumCreditScore": 300,
"maximumCreditScore": 700,
"maximumAmount": 10000
}
}, {
"auto-loan": {
"id": 2,
"provider": "Blue Harbor",
"term": 24,
"minimumCreditScore": 600,
"maximumCreditScore": 700,
"maximumAmount": 5000
}
}]

Consider that the driver with one trip want to pick up some peoples in different locations like this:
String[] locations ={
"person1, person2, person3, person4, person5",
" person6, person7, person8, person9",
"person10, person11, person12",
"person13, person14, person15",}
in each location there are different choice, so write a code present all possible way to pick up people in the different locations. you can use every data structure needs.

given preorder traversal [5,3,2,4,8,7,9] of a BST, how do we identify the leaf nodes without building the tree ?

@Anonymous Thanks for the reply!
Please try other use cases like, only single leaf node

There is dedicated Samsung software for coding test the question is given below:
There is one spaceship. X and Y co-odinate of source of spaceship and destination spaceship is given. There are N number of warmholes each warmhole has 5 values.
First 2 values are starting co-ordinate of warmhole and after that value no. 3 and 4 represents ending co-ordinate of warmhole and last 5th value is represents cost to pass through this warmhole. Now these warmholes are bi-direction.
Now the to go from (x1,y1) to (x2,y2) is abs(x1-x2)+abs(y1-y2).
The main problem here is to find minimum distance to reach spaceship from source to destination co-ordinate using any number of warm-hole. It is ok if you wont use any warmhole.

list1 -->aaa,bbb,ddd,xyxz,...
list2-->bbb,ccc,ccc,hkp,..
list3> ddd,eee,,ffff,lmn,..

Inside a list the words are sorted
I want to remove words which are repeated across the list and print in sorted order
If the words are repeated in same list its valid.
In the above case
it should print aaa-->ccc-->ccc-->eee--->fff-->glk-->hkp-->lmn-->xyxz

Some questions about how to write a immutable class.

Find top n cities which got most orders. For example, amazon got a list of orders, and these orders will be shipped to different cities.

# 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.
# 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

I recently had a take home assignment for a optimization engineer position at FB. I got 24 hours to finish it however it was suggested to finish under 90 minutes. The problem was to calculate the dot product of sparse matrix (optimize for speed). I wrote following code but got dinged. Not sure what's wrong to not clear even first stage!

#include <iostream>
#include <string>
#include <map>
#include <sstream>
using namespace std;
class SparseVector{
private:
    map<int, double> m_map;
    int m_size;
public:
    SparseVector(int size){
        this->m_size = size;
        this->m_map =  *new map<int, double>();
    }
    void setValue(int i, double value){
        if( i < 0 || i > m_size)
            return;
        if(value == 0.0){
            map<int, double>::iterator it = m_map.find(i);
            if(it != m_map.end())
                m_map.erase(it);
        }
        else
            m_map[i] = value;
    }
    double getValue(int i){
        if( i < 0 || i > m_size)
            return 0.0;
        else {
            map<int, double>::iterator it = m_map.find(i);
            if(it != m_map.end())
                return it->second;
            else
                return 0.0;
        }
    }
    int getSize(){
        return m_size;
    }
    
    static double s_dotProduct(SparseVector a, SparseVector b){
        if (a.m_size != b.m_size)
            return 0.0;
        
        double sum = 0.0;
        if (a.m_map.size() <= b.m_map.size())
        {
            for (map<int, double>::iterator it = a.m_map.begin() ; it != a.m_map.end(); it++)
                if ((b.m_map.find(it->first)) != b.m_map.end())
                    sum += a.getValue(it->first) * b.getValue(it->first);
        }
        else
        {
            for (map<int, double>::iterator it = b.m_map.begin() ; it != b.m_map.end(); it++)
                if ((a.m_map.find(it->first)) != a.m_map.end())
                    sum += a.getValue(it->first) * b.getValue(it->first);
        }
        return sum;
    }
    template <typename T>
    static string s_ToString(T val)
    {
        stringstream stream;
        stream << val;
        return stream.str();
    }
    static string s_printSparseVector(SparseVector a)
    {
        string s = "";
        for (map<int, double>::iterator it =  a.m_map.begin(); it!= a.m_map.end(); it++)
            s += "(" +  s_ToString(it->first) + ", " + s_ToString(it->second) + ") ";
        return s;
    }
};

int main(int argc, char *argv[]){
     int lenA = 0, lenB = 0, setValA = 0, setValB = 0;
     cout << "Enter length for Sparse vector A : " << endl;
     cin>> lenA;
     cout << "Enter length for Sparse vector B : " << endl;
     cin>> lenB;
     
     SparseVector v1 = SparseVector(lenA);
     SparseVector v2 = SparseVector(lenB);
     double y; int x;
     
     cout << "Enter number of set values for Sparse vector A : " << endl;
     cin >> setValA;
     cout << "Enter the set Index, Value pair in separate lines" << endl;
     for(int i = 0; i < setValA; i++){
     cin >> x;
     cin >> y;
     v1.setValue(x, y);
     }
     
     cout << "Enter number of set values for Sparse vector B : " << endl;
     cin >> setValB;
     cout << "Enter the set Index, Value pair in separate lines" << endl;
     for(int j = 0; j < setValB; j++){
     cin >> x;
     cin >> y;
     v2.setValue(x, y);
     }
cout <<"Sparse Vector1 = " << SparseVector::s_printSparseVector(vectorA) << endl ;
     cout <<"Sparse Vector2 = " << SparseVector::s_printSparseVector(vectorB) << endl;
     cout <<"Dot product of two sparse vectors is  = " <<  SparseVector::s_dotProduct(vectorA, vectorB) << endl;
     return 0;
}

You are given a series of moves as a string, such as ">^V<^". You need to output a list with all the series of moves that represent a loop if they were made by a robot for example.

A loop is defined as a series of moves that lead to the same position, for example "><" or "^V".

In the case of "^>V<^" the output should be ["^>V<", "^>V<"], essentially the last move generates a new loop because it leads to a position that was visited before and that naturally can be followed up to the same position and describe a loop.

Giving a string and an string dictionary, find the longest string in dictionary which can formed by deleting some characters of the giving string.
eg:S = abpcplea, Dict = {ale, apple, monkey, plea}, the return "apple"

I was thinking of the following approach,

Build a Trie for all words in the Dict - O( n * k) where k is the longest string in the dict
For each character c, in S check if there is a word in Trie that starts with it and has the letters that appear in S after c. We can short circuit based on remaining characters and the length of longest string found so far.
This should take O( N * k) where N is length of S.

Find all comments in the Java (it could be Python or any other language of your choice) codes that’s parsed in as a string.
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”
]

Q: If you were given a series of equations e.g. [A = B, B = D, C = D, F = G, E = H, H = C]
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.

Given an input string "aabbccba", find the shortest substring from the alphabet "abc".

In the above example, there are these substrings "aabbc", "aabbcc", "ccba" and "cba". However the shortest substring that contains all the characters in the alphabet is "cba", so "cba" must be the output.

Output doesnt need to maintain the ordering as in the alphabet.

Other examples:
input = "abbcac", alphabet="abc" Output : shortest substring = "bca".

Given an array of integers, design an algorithm that moves all non-zero integers to the end of the array. Minimize the number of writes or swaps.

Write code to decode strings. For example, String str = "3[a2[bd]g4[ef]h]", the output should be "abdbdgefefefefhabdbdgefefefefhabdbdgefefefefh".

My solution is as follows.

public class StringDecoder {
	
	public static void main(String[] args){
		
		String s = "3[a2[bd]g4[ef]h]";
		
		System.out.println(decode(s));
		
	}
	
	public static String decode(String s){
		
		if(s == null || s.length()==0) return s;
		
		int indexOfFirstNumber = findIndexOfFirstNumber(s);
		int indexOfFirstBracket = findIndexOfFirstBracket(s);
		int indexOfClosingBracket = findIndexOfClosingBracket(s, indexOfFirstBracket);
		
		if(indexOfFirstNumber == -1) return s;
		
		String subStr1 = s.substring(0, indexOfFirstNumber);
		String subStr2 = decode(s.substring(indexOfFirstBracket+1, indexOfClosingBracket));
		String subStr3 = decode(s.substring(indexOfClosingBracket+1, s.length()));
		
		int duplicates = Integer.parseInt(s.substring(indexOfFirstNumber, indexOfFirstBracket));
		
		StringBuilder sb = new StringBuilder();
		sb.append(subStr1);
		
		while(duplicates>0){
			sb.append(subStr2);
			duplicates--;
		}
		
		sb.append(subStr3);
		
		return sb.toString();
	}
	

	public static int findIndexOfFirstNumber(String s){
		
		int index = -1;
		for(int i=0; i<s.length(); i++){
			
			char c = s.charAt(i);
			if(c>=47 && c<=58){
				index = i;
				break;
			}
			
		}
		
		return index;
	}
	
	public static int findIndexOfFirstBracket(String s){
		
		int index = -1;
		for(int i=0; i<s.length(); i++){
			
			char c = s.charAt(i);
			if(c=='['){
				
				index = i;
				break;
			}
			
		}
		
		return index;
	}
	
	public static int findIndexOfClosingBracket(String s, int indexOfBracket){
		int index = -1;
		
		int numberOfBracket = 1;
		
		for(int i=indexOfBracket+1; i<s.length(); i++){
			
			char c = s.charAt(i);
			if(c == '[') numberOfBracket++;
			
			if(c==']'){
				numberOfBracket--;
				if(numberOfBracket==0){
					index = i;
					break;
				}
			}
		}
		
		return index;
	}
}

Given 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.

e.g.
a relies on b
b relies on c

then a valid sequence is [c, b, a]