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

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++ code for Maximum size square  
// sub-matrix with all 1s  
#include <bits/stdc++.h> 
#define bool int  
#define R 6  
#define C 5  
using namespace std; 
  
  
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],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;  
            }  
        }              
    }  
  
    cout<<"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--)  
        {  
            cout << M[i][j] << " ";  
        }  
        cout << "\n";  
    }  
}  
  
  
/* Driver code */
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);  
}  
  
// This is code is contributed by rathbhupendra 

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); 
    } 
  
} 

PHP

<?php 
// PHP code for Maximum size square  
// sub-matrix with all 1s  
  
function printMaxSubSquare($M, $R, $C)  
{  
    $S = array(array()) ; 
  
    /* 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--)  
        {  
            echo $M[$i][$j], " " ;  
        }  
        echo "\n" ; 
    }  
}  
  
# Driver code 
$M = array(array(0, 1, 1, 0, 1),  
           array(1, 1, 0, 1, 0),  
           array(0, 1, 1, 1, 0),  
           array(1, 1, 1, 1, 0),  
           array(1, 1, 1, 1, 1),  
           array(0, 0, 0, 0, 0));  
      
$R = 6 ; 
$C = 5 ;          
printMaxSubSquare($M, $R, $C);  
  
// This code is contributed by Ryuga 
?> 

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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My Personal Notes

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

C++

C

Java

Python3

C#

PHP

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