In last post, we saw a dynamic programming approach to for finding maximum size square sub-matrix with all 1s. In this post, we will discuss how to find largest all 1s sub-matrix in a binary matrix. The resultant sub-matrix is not necessarily a square sub-matrix.**Example:**

Largest all 1s sub-matrix is from (1,1) to (2,4).1 1 0 0 1 00 1 1 1 1 11 1 1 1 1 00 0 1 1 0 0

1 1 1 11 1 1 1

**Algorithm:**If we draw a histogram of all 1’s cells in above rows (until we find a 0) for a particular row, then maximum all 1s sub-matrix ending in that row will the equal to maximum area rectangle in that histogram. Below is the example for 3rd row in above discussed matrix:

For above purpose, we need to generate an auxiliary matrix S[][] where each element represents the number of 1s above and including it, up until the first 0. S[][] for above matrix will be as shown below:

Now we can simple call our maximum rectangle in histogram on every row in S[][] and update the maximum area every time.1 1 0 0 1 00 2 1 1 2 11 3 2 2 3 00 0 3 3 0 0

Also we don’t need any extra space for saving S. We can update original matrix (A) to S and after calculation, we can convert S back to A.

**Code:**I am not writing the code for largestArea() function. One can find its definition in this post.

#define ROW 10

#define COL 10

int find_max_matrix(int A[ROW][COL])

{

int i, j;

int max, cur_max;

cur_max = 0;

//Calculate Auxilary matrix

for (i=1; i<ROW; i++)

for(j=0; j<COL; j++)

{

if(A[i][j] == 1)

A[i][j] = A[i-1][j] + 1;

}

//Calculate maximum area in S for each row

for (i=0; i<ROW; i++)

{

max = largestArea(A[i], COL);

if (max > cur_max)

cur_max = max;

}

//Regenerate Oriignal matrix

for (i=ROW-1; i>0; i--)

for(j=0; j<COL; j++)

{

if(A[i][j])

A[i][j] = A[i][j] - A[i-1][j];

}

return cur_max;

}

**Complexity:**Lets say that total number of rows and columns in A are m and n respectively and N = m*n

Complexity of calculating S = O(m*n) = O(N)

Complexity of LargestArea() for every row = O(n) since there are n elements in every row. Since we called LargestArea() m times so total complexity of calculating largest area = O(m*n) = O(N)

Complexity of converting S to A = O(m*n) = O(N)

So total time complexity = O(N)

Since we are not using any extra space, so space complexity is O(1).

**Note:**Above function only returns largest area, which can be very easily modified to get start and row indexes as well.

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