# finding the maximum number of 1's in a row

A 2d matrix is given filled with 1's and 0's. It is given that all 1's in a row come before all the 0's. We have to find the maximum number of 1's in a row.

I have made the solution that we can apply binary search on every row to get the last index of last 1 in that row before 0's begin and hence the no. of 1's will be its index+1. So we can do this at every row. So the complexity would be O(mlogn),where m is the no. of rows and n is the no. of columns. Can there be a better solution to this?

## Answers

It can be done in O(n+m).

Start with curmax equal to 0.

Then process rows one by one. Increase curmax while there are at least curmax ones in that row, i.e. checking if curmax value is one.

The answer will be curmax-th, after all rows are processed.

This will work in O(n+m).

Will it be quicker than O(m*logn)? It depends. If m is less then n/(log(n) - 1) it will work, actually longer then O(m*log n) and quicker otherwise, just in terms of complexity. Considering constants is another problem, when approximating time. So for n and m of one magnitude this will be quicker, for different there is only one choice - try both, pick better.