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.

 
   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

The maximum square sub-matrix with all set bits is

    1  1  1
    1  1  1
    1  1  1

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.

#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("\n 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();  }

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