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.
maximum-size-square-sub-matrix-with-all-1s



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);
    }
  
}

Python3

# Python3 code for Maximum size
# square sub-matrix with all 1s
  
def printMaxSubSquare(M):
    R = len(M) # no. of rows in M[][]
    C = len(M[0]) # no. of columns in M[][]
  
    S = [[0 for k in range(C)] for l in range(R)]
    # here we have set the first row and column of S[][]
  
    # Construct other entries
    for i in range(1, R):
        for j in range(1, C):
            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
    # indices of maximum entry in S[][]
    max_of_s = S[0][0]
    max_i = 0
    max_j = 0
    for i in range(R):
        for j in range(C):
            if (max_of_s < S[i][j]):
                max_of_s = S[i][j]
                max_i = i
                max_j = j
  
    print("Maximum size sub-matrix is: ")
    for i in range(max_i, max_i - max_of_s, -1):
        for j in range(max_j, max_j - max_of_s, -1):
            print (M[i][j], end = " ")
        print("")
  
# Driver Program
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)
  
# This code is contributed by Soumen Ghosh

C#

// C# Code for Maximum size square 
// sub-matrix with all 1s
  
using System;
  
  
public class GFG
{
    // method for Maximum size square sub-matrix with all 1s
    static void printMaxSubSquare(int [,]M)
    {
        int i,j;
        //no of rows in M[,]
        int R = M.GetLength(0); 
         //no of columns in M[,]
        int C = M.GetLength(1); 
        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;
                }    
            }                
        }    
          
        Console.WriteLine("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--)
            {
                Console.Write(M[i,j] + " ");
            
            Console.WriteLine();
        
    
      
    // Driver program 
    public static void Main() 
    {
        int [,]M = new int[6,5]{{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



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Improved By : tufan_gupta2000