**Question: **Given a matrix consisting only 0s and 1s, find the maximum size square sub-matrix with all 1s.

**Example:** Consider the below matrix.

The maximum square sub-matrix with all '1' bits is from (2,1) to (4,3)0 1 1 0 11 1 0 1 00 1 1 1 01 1 1 1 01 1 1 1 10 0 0 0 0

1 1 11 1 11 1 1

**Answer:**This is a classic Dynamic Programming problem. Lets calculate the maximum size square sub-matrix as we traverse the original matrix M[][]. We will use a auxiliary matrix S[][] of same size for memoization. S[i][j] represents size of the square sub-matrix with all 1s including M[i][j]. 'i' and 'j' will be the last row and column respectively in square sub-matrix.

**How to calculate S[i][j]:**

We should note that if M[i][j] is '0' then S[i][j] will obviously be '0'. If M[i][j] is '1' then S[i][j] depends on earlier values.

If M[i][j] is '1' then it will contribute to the all 1s square sub-matrix ending at either M[i][j-1] or M[i-1][j] or M[i-1][j-1]. If we visualize the conditions then, we will see:

S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1

**How did we arrive at above relationship?**

Note if we include M[i][j] in earlier calculated sub-matrix then we are adding S[i][j] elements from i

^{th}row and j

^{th}columns. They all should be '1' if we wanna include M[i][j]. On visualizing with some examples, readers will analyze why, minimum of 3 neighbors is taken.

For the given M[][] in above example, constructed S[][] would be:

The value of maximum entry in above matrix is 3 and coordinates of the entry are (4, 3). Using the maximum value and its coordinates, we can find out the required sub-matrix.0 1 1 0 11 1 0 1 00 1 1 1 01 1 2 2 01 2 2 3 10 0 0 0 0

**Code:**

#define ROW 10

#define COL 10

void FindMaxSubSquare(bool M[ROW][COL], int &max_i, int &max_j, int &size)

{

int i,j;

int S[ROW][COL];

/* Set first column of S[][]*/

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

S[i][0] = M[i][0];

/* Set first row of S[][]*/

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

S[0][j] = M[0][j];

/* Construct other entries of S[][]*/

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

{

for(j = 1; j < COL; j++)

{

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

S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1;

else

S[i][j] = 0;

}

}

/* Find the maximum entry, and indexes of maximum entry in S[][] */

size = S[0][0]; max_i = 0; max_j = 0;

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

{

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

{

if(size < S[i][j])

{

size = S[i][j];

max_i = i;

max_j = j;

}

}

}

return

}

**Complexity:**

Time Complexity: O(m*n) where m is number of rows and n is number of columns in the given matrix.

Space Complexity: O(m*n) where m is number of rows and n is number of columns in the given matrix.

**Note:**Part of this post is taken from geeksforgeeks.

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