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I forgot my Sage Drive password, how do I reset it?

What is Sage Drive?

The Connection Manager could not start the database engine In Sage ?

The company file is on the server or another computer.?

0x80040702 "Failed to load dll: Reg 32" when installing Sage ?

How to Disable Kodak Printer Ink Monitor ?

How do I clean the printhead on my Kodak ESP c310?

How do I calibrate my Kodak printer?

How do I reset my Xerox WorkCentre 7835?

Given two strings, A and B, of equal length, find whether it is possible to cut both strings at a common point such that the first part of A and the second part of B form a palindrome.

Extension1. How would you change your solution if the strings could be cut at any point (not just a common point)?

Extension2. Multiple cuts in the strings (substrings to form a palindrome)? Form a palindrome using a substring from both strings. What is its time complexity?

You need to test a time converter that converts minutes to seconds, hours, days, weeks, months, years. What kind of test cases would you recommend? We enter only minutes, and the site automatically translates them into seconds, hours, days, weeks, months, years. This is how it looks:

[0]minutes
----------------
[0] seconds
[0] hours
[0] days
[0] weeks
[0] months
[0] years
Values change dynamically, when input data changes - minutes

/*
## Setup

The flow of a dispute is as follows:
- A charge is created by an end customer.
- Stripe receives a dispute record from the bank.
- The business responds with evidence.
- If no second dispute is received within 30 days after evidence submission, the dispute is won. If a second dispute is received, the dispute is lost.

Charge
(Maybe) Dispute Record
(Maybe) Evidence submission
(Maybe) Second Dispute Record

The raw tables generated from the API look like:

```
Charges
+---------------+-----------+
| charge_id | varchar |
| created | timestamp |
| amount | int |
| seller_id | varchar |
| customer_id | varchar |
+---------------|-----------+

Dispute Records
+----------------+-----------+
| dispute_id | varchar |
| created | timestamp |
| charge_id | varchar |
+----------------|-----------+

Evidence Submission
+-------------------+-----------+
| evidence_id | varchar |
| created | timestamp |
| charge_id | varchar |
+-------------------|-----------+
```
*/

/*

1. Can you design a unified dispute table that would allow us to compute things like the win rate, dispute rate, evidence submission rate etc?

*/

A happy number is defined by the following process. Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers, while those that do not end in 1 are unhappy numbers.
Input:
Your program should read lines of text from standard input. Each line contains a single positive integer, N.
Output:
If the number is a happy number, print 1 to standard output. Otherwise, print 0.

Given two binary trees, explain how you would create a diff such that if you have that diff and either of the trees you should be able to generate the other binary tree. Implement a function which takes Node tree1, Node tree2 and returns that diff.

C program for the given two array as point of x and y
int x[] = { 2,3,2,4,2};
int y[] = {2, 2, 6,5,8}; count the pair
maximum area coverd in x y plane
Ouput:3
From the above input there are five coordinates possible (2,2)(3,2)(2,6()(4,5)and (2,8) in which (2,2)(2,6)(2,8)that is three coodinaters covers the maximum area in x y plane
Input: x[] ={1,2,3}
Y[] = {1,2,3}
Output: 1
there are three coordinates (1,1)(2,2)(3,3)in which only one coordinates covers maximum distancethat is (1,3)

Create an iterator class that stores a list of the built-in Iterators.
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

It will gain you more knowledge, intensify your soft skills, strong work ethics and grow your network. What is it?

Hi,

could you please advise how to land a job in Microsoft. I have applied for many openings through LinkedIn but not getting contacted from Microsoft recruiters or have received any updates yet.

Thanks
Deepa

Program to print last element of matrix by traversing a matrix by anticlockwise starting from cell(1,1) and skiping alternate.
Ex:
Input:1 2 3 4
2 4 5 1
7 9 6 2
3 5 2 7
Output: 5
Start traversing matrix from cell(1,1) skip next row element go to 3 row element again skip 4 element go to bottom of second column .....up to last element and print only last element.
If Input: 1 2
3 4
Output: 2
Input:1 2 3
4 5 6
7 8 9
Output: 5

Find whether string S is periodic.
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"

Given string S, find out the P (repetitive pattern) of S.

Given a dictionary of words & a miss-spelled input, write a function which will find 3 words from the dictionary which are closest (by difference of 1-character) to the given input.

eg - dict = {vil, sit, flick, pat, pluck, sat, vat}, input = vit, ans = {sit, vil, vat}

Print all possible combination by replacing special charecters from number.

// For a given vector of integers and integer K, find the number of non-empty subsets S such that min(S) + max(S) <= K
// For example, for K = 8 and vector [2, 4, 5, 7], the solution is 5 ([2], [4], [2, 4], [2, 4, 5], [2, 5])
The time complexity should be O(n2). Approach and code was asked

Find the indices of all anagrams of a given word in a another word.
For example: Find the indices of all the anagrams of AB in ABCDBACDAB (Answer: 0, 4, 8)

I am given an array of Transactions in a ledger. A Transaction object has three things.
1. Sender (which means who started this transaction)
2. Receiver (means who is the destination of transaction)
3. Timestamp (at what time this transaction was executed)

Now I need to write a method findIfTransactionIsValid() which will have the array of all transactions, one sender, one receiver. A transaction is valid is following cases:
1. if sender and receiver are same
2. The timestamp should be increasing (what I mean here is if A -> B happens at Time 2 and B-> C happens at Time 1, then A->C is not a valid transaction, however if B -> C happens at Time 3 then A--> C is a valid transaction)

Example:
Transaction
{
Sender;
Receiver;
Timestamp;
}

Example: [T is timestamp here]
A -> B (T=0)
B -> C (T=1)
C -> F (T=0)

findIfTransactionIsValid(A, C) -> this should return true
findIfTransactionIsValid(B,F) -> false (time is backwards)

If the question is still not clear, please see the solution I wrote. I have written a recursive solution and is working but I am seeing help to improve the solution.

Write a retry function, continue to fetch data until u have exhausted max entries. If it fails, continue to retry until retry's have been exhausted.

There are N countries, each country has Ai players. You need to form teams of size K such that each player in the team is from a different country.
Given N and number of players from each country and size K. Find the maximum number of teams you can form.

Given 2 trees T1 & T2 (both can have > 2 childs), write an algorithm to find if T2 is a subtree of T1.

Follow up question, for any branch in T1
a->b->c->d

the following is a valid branch in tree T2(i.e. the isSubTree() algorithm mush evaluate to true in below circumstances)

a->d
a->c->d
c->d

You are given a 2d grid where each grid item has a value of 1 or 0, you can only move horizontally or vertically and if both blocks have value of 1. You are also given a starting index, the output should have the "connected" grid items property to true.

For example:

input = [
      [{value: 0}, {value: 1}, {value: 1}],
      [{value: 0}, {value: 0}, {value: 1}],
      [{value: 1}, {value: 1}, {value: 1}]
    ];

    startRowIndex = 2;
    startColumnIndex = 0;
    
    output = [
      [{value: 0}, {value: 1, connected: true}, {value: 1, connected: true}],
      [{value: 0}, {value: 0}, {value: 1, connected: true}],
      [{value: 1, connected: true}, {value: 1, connected: true}, {value: 1, connected: true}]
    ];

This is the first part of the question, this can be easily solved using either DFS or BFS.

The second part is you are given the output of the first function and the same start indices. Along with these two input arguments, you are also given a flipIndex. The grid item at the given flip index will have the value flipped. Now give the updated matrix with the updated "connected" path.

input = [
      [{value: 0}, {value: 1, connected: true}, {value: 1, connected: true}],
      [{value: 0}, {value: 0}, {value: 1, connected: true}],
      [{value: 1, connected: true}, {value: 1, connected: true}, {value: 1, connected: true}]
    ];

    startRowIndex = 2;
    startColumnIndex = 0;

    flipRowIndex = 1;
    flipColumnIndex = 2;
    
    output = [
      [{value: 0}, {value: 1}, {value: 1}],
      [{value: 0}, {value: 0}, {value: 0}],
      [{value: 1, connected: true}, {value: 1, connected: true}, {value: 1, connected: true}]
    ];

Optimize the runtime of the following function from O(I * R) to O(I + R):

const cars = [
  {make: "toyota", color: "red", cat: "suv"},
  {make: "honda", color: "black", cat: "sedan"},
  {make: "toyota", color: "red", cat: "sedan"},
  {make: "nissan", color: "blue", cat: "suv"},
];

const rejects = [
  {key: "color", value: "black"},
  {key: "color", value: "blue"},
  {key: "make", value: "honda"},
];

function rejectByKeyValues(items, rejectKeyValues) {
  rejectKeyValues.forEach(keyValue => {
    items = items.filter(ii => ii[keyValue.key] !== keyValue.value);
  });
  return items;
}