SDE - 1

Step 1: Since this is a Dynamic Programming problem So the first approach I have explained is the recursive approach.In which we can break the problem into subproblems.
Step 2: It was not working for all test cases.
Step 3:So I tried to optimize it with Memoization. 
Step 4: It worked for all Test Cases.

Step 1: Since it is the Greedy Problem So firstly I created a vector of 3 items start time, end time, and meeting number, and sorted this using comparator based on ascending order of end time.

Step 2: We need a variable to store the answer. Initially, the answer is 1 because the first meeting can always be performed. Make another variable, say limit that keeps track of the ending time of the meeting that was last performed. Initially set limit as the end time of the first meeting.

Start iterating from the second meeting. At every position we have two possibilities:-

If the start time of a meeting is strictly greater than the limit we can perform the meeting. Update the answer. Our new limit is the ending time of the current meeting since it was last performed. Also update the limit. 
If the start time is less than or equal to the limit, skip and move ahead.
This way this code was passing all test cases

Step1: I first told him the brute force which is direct for this problem, but of course, he was not satisfied.
Step 2: We map the indices of matrix elements to a 1D array as follows : 
If we copy the elements of a matrix (say ‘MAT’) to an array (say ‘ARR’) in a row-wise manner, then the element ‘MAT[i][j]' is equal to ‘ARR[N*i+j]’, where ‘N’ denotes the number of columns in ‘MAT’.
Similarly, converting ‘ARR’ back to ‘MAT’, the element ‘ARR[i]’ is equal to ‘MAT[i/N][i%N]’ as ‘ARR’ contains M*N elements, so ‘rowNum’ is equal to i/N and ‘column' is equal to i%N.
Step3: After this approach he was satisfied.

Tip 1 : In this round communication is very important.
Tip 2 : Doing the top 20 HR Questions from any website is sufficient for this round.