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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}]
    ];