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++



filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


C

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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(); 

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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);
    }
  
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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);
    }
  
}

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right



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

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.




My Personal Notes arrow_drop_up