For a_i, j we calculate p_i, j which is position of next 0 element in a row. If a_i, j == 0, p_i, j == j. If there is no 0 next in a row, p_i, j = m.

For example for matrix

0 1 0

1 1 1

0 0 0

We get

0 2 2

3 3 3

0 1 2

Now for each column j sort all p_i, j - a_i, j and count cnt_w — number of elements which is greater than w for w in 1..m. Answer is max(w * cnt_w). It's solution for O(n² * log_n)

Build a dp array where dp[i][j] denotes maximum no of consecutive ones in row i, starting at column index j.

Now sort the dp values for each column, and rest part is easy. It works in O(n²log(n))