Software Engineer

I firstly thought of applying brute force by created one array of size n * m and put sum of element picking one from array 1 and then other from array 2 and sorted that array and kept track of which number is sum of which pair.
but this had very high time complexity: O(n * m * log (n * m)).
Then I went ahead to think a little bit more different than Brute force so I came up with below approach:
I tried finding k smallest sum pairs, starting from least sum pair. The idea was to keep track of all elements of one array and which have been already considered for every element of other array so that in an iteration we only consider next element. For this purpose, we use an index array index2[] to track the indexes of next elements in the other array. It simply means that which element of second array to be added with the element of first array in each and every iteration. We increment value in index array for the element that forms next minimum value pair.

By visualising the problem using graph with x axis as time and y axis as trains so the answer for minimum platforms would be maximum number of trains in same time interval.
And then going into solution:
Using two loops, take one interval and check it with other intervals. Keep updating the maximum in each outer loop iteration.
But this is having time complexity as O(n^2)
So Interviewer asked to look at improving the time complexity.
So I thought for sometime and came up with keeping track of trains which has arrived but did not depart.
So for this, we have to consider events in sorted order.
This is taking time O(n * log n)

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