Find index of the row containing maximum number of 1’s in a binary matrix

Given a binary M x N row-wise sorted matrix, find a row which contains maximum number of 1 in linear time.

For example,


[ 0 0 0 1 1 ]
[ 0 0 1 1 1 ]
[ 0 0 0 0 0 ]
[ 0 1 1 1 1 ]
[ 0 0 0 0 1 ]

Output: Maximum 1’s are present in the row 4


The idea is to start from the top-right corner of the matrix and do the following

– if current cell has value 1, continue moving left till we encounter 0 or all columns are processed, else
– if current cell has value 0, continue moving down till we encounter 1 or all rows are processed.

Finally, we return the row index of last cell in which we have seen 1.

C++ implementation –

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Maximum 1’s are present in the row 4


The time complexity of above solution is O(M + N) and auxiliary space used by the program is O(1).

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what is the logic behind this? How does it work correctly?


ok got it. didn’t read the problem carefully “row wise sorted binary matrix”.


I dont think this is only move down if you hit a 0 in the very last element in every row.. therefore you get 2. I tried the exact code and it didn’t work. Second of all, how is the complexity O(N+M) ? You are going through the matrix row-wise and column-wise. I think it’s O(N*M).


Code works fine. I think you’re confused somewhere. Please read the problem statement again and try to relate the logic. If you still think you’re right, kindly provide a test case where the code fails.
Also time complexity remains O(N+M) since we either go left or bottom, and not in other two directions at any point.


Sorry, just made the same mistake as arandomguy below. The solution is correct.