# Maximum size square sub-matrix with all 1s

Given a binary matrix, find out the maximum size square sub-matrix with all 1s.

For example, consider the below binary matrix.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Algorithm:
Let the given binary matrix be M[R][C]. The idea of the algorithm is to construct an auxiliary size matrix S[][] in which each entry S[i][j] represents size of the square sub-matrix with all 1s including M[i][j] where M[i][j] is the rightmost and bottommost entry in sub-matrix.

```1) Construct a sum matrix S[R][C] for the given M[R][C].
a)    Copy first row and first columns as it is from M[][] to S[][]
b)    For other entries, use following expressions to construct S[][]
If M[i][j] is 1 then
S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1
Else /*If M[i][j] is 0*/
S[i][j] = 0
2) Find the maximum entry in S[R][C]
3) Using the value and coordinates of maximum entry in S[i], print
sub-matrix of M[][]```

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

```   0  1  1  0  1
1  1  0  1  0
0  1  1  1  0
1  1  2  2  0
1  2  2  3  1
0  0  0  0  0```

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.

## C/C++

```// C/C++ code for Maximum size square sub-matrix with all 1s
#include<stdio.h>
#define bool int
#define R 6
#define C 5

void printMaxSubSquare(bool M[R][C])
{
int i,j;
int S[R][C];
int max_of_s, max_i, max_j;

/* Set first column of S[][]*/
for(i = 0; i < R; i++)
S[i][0] = M[i][0];

/* Set first row of S[][]*/
for(j = 0; j < C; j++)
S[0][j] = M[0][j];

/* Construct other entries of S[][]*/
for(i = 1; i < R; i++)
{
for(j = 1; j < C; 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[][] */
max_of_s = S[0][0]; max_i = 0; max_j = 0;
for(i = 0; i < R; i++)
{
for(j = 0; j < C; j++)
{
if(max_of_s < S[i][j])
{
max_of_s = S[i][j];
max_i = i;
max_j = j;
}
}
}

printf("Maximum size sub-matrix is: \n");
for(i = max_i; i > max_i - max_of_s; i--)
{
for(j = max_j; j > max_j - max_of_s; j--)
{
printf("%d ", M[i][j]);
}
printf("\n");
}
}

/* UTILITY FUNCTIONS */
/* Function to get minimum of three values */
int min(int a, int b, int c)
{
int m = a;
if (m > b)
m = b;
if (m > c)
m = c;
return m;
}

/* Driver function to test above functions */
int main()
{
bool M[R][C] =  {{0, 1, 1, 0, 1},
{1, 1, 0, 1, 0},
{0, 1, 1, 1, 0},
{1, 1, 1, 1, 0},
{1, 1, 1, 1, 1},
{0, 0, 0, 0, 0}};

printMaxSubSquare(M);
getchar();
}
```

## Java

```// JAVA Code for Maximum size square sub-matrix with all 1s
public class GFG
{
// method for Maximum size square sub-matrix with all 1s
static void printMaxSubSquare(int M[][])
{
int i,j;
int R = M.length;         //no of rows in M[][]
int C = M[0].length;     //no of columns in M[][]
int S[][] = new int[R][C];

int max_of_s, max_i, max_j;

/* Set first column of S[][]*/
for(i = 0; i < R; i++)
S[i][0] = M[i][0];

/* Set first row of S[][]*/
for(j = 0; j < C; j++)
S[0][j] = M[0][j];

/* Construct other entries of S[][]*/
for(i = 1; i < R; i++)
{
for(j = 1; j < C; j++)
{
if(M[i][j] == 1)
S[i][j] = Math.min(S[i][j-1],Math.min(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[][] */
max_of_s = S[0][0]; max_i = 0; max_j = 0;
for(i = 0; i < R; i++)
{
for(j = 0; j < C; j++)
{
if(max_of_s < S[i][j])
{
max_of_s = S[i][j];
max_i = i;
max_j = j;
}
}
}

System.out.println("Maximum size sub-matrix is: ");
for(i = max_i; i > max_i - max_of_s; i--)
{
for(j = max_j; j > max_j - max_of_s; j--)
{
System.out.print(M[i][j] + " ");
}
System.out.println();
}
}

// Driver program
public static void main(String[] args)
{
int M[][] =  {{0, 1, 1, 0, 1},
{1, 1, 0, 1, 0},
{0, 1, 1, 1, 0},
{1, 1, 1, 1, 0},
{1, 1, 1, 1, 1},
{0, 0, 0, 0, 0}};

printMaxSubSquare(M);
}

}
```

Output:

```Maximum size sub-matrix is:
1 1 1
1 1 1
1 1 1
```

Time Complexity: O(m*n) where m is number of rows and n is number of columns in the given matrix.
Auxiliary Space: O(m*n) where m is number of rows and n is number of columns in the given matrix.
Algorithmic Paradigm: Dynamic Programming

Please write comments if you find any bug in above code/algorithm, or find other ways to solve the same problem

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3.3 Average Difficulty : 3.3/5.0
Based on 256 vote(s)

Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.