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Maximum size square sub-matrix with all 1s

  • Difficulty Level : Medium
  • Last Updated : 12 May, 2022

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 the size of the square sub-matrix with all 1s including M[i][j] where M[i][j] is the rightmost and bottom-most 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 the 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 the maximum entry in the above matrix is 3 and the 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], S[i-1][j],
                                S[i-1][j-1]}) + 1; //better of using min in case of arguments more than 2
            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 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 = []
    for i in range(R):
      temp = []
      for j in range(C):
        if i==0 or j==0:
          temp += M[i][j],
        else:
          temp += 0,
      S += temp,
    # here we have set the first row and first column of S same as input matrix, other entries are set to 0
 
    # Update 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
?>

Javascript




<script>
// JavaScript code for Maximum size square
// sub-matrix with all 1s
let R = 6;
let C = 5;
 
function printMaxSubSquare(M) {
    let i,j;
    let S = [];
 
for ( var y = 0; y < R; y++ ) {
    S[ y ] = [];
    for ( var x = 0; x < C; x++ ) {
        S[ y ][ x ] = 0;
    }
}
    let 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;
            }
        }            
    }
 
    document.write("Maximum size sub-matrix is: <br>");
    for(i = max_i; i > max_i - max_of_s; i--)
    {
        for(j = max_j; j > max_j - max_of_s; j--)
        {
            document.write( M[i][j] , " ");
        }
        document.write("<br>");
    }
}
 
 
/* Driver code */
let 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);
</script>

Output: 

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

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

Space Optimized Solution: In order to compute an entry at any position in the matrix we only need the current row and the previous row.

C++




// C++ code for Maximum size square
// sub-matrix with all 1s
// (space optimized solution)
#include <bits/stdc++.h>
 
using namespace std;
 
#define R 6
#define C 5
 
void printMaxSubSquare(bool M[R][C])
{
    int S[2][C], Max = 0;
   
    // set all elements of S to 0 first
    memset(S, 0, sizeof(S));
 
    // Construct the entries
    for (int i = 0; i < R;i++)
        for (int j = 0; j < C;j++){
 
            // Compute the entrie at the current position
            int Entrie = M[i][j];
            if(Entrie)
                if(j)
                    Entrie = 1 + min(S[1][j - 1], min(S[0][j - 1], S[1][j]));
 
            // Save the last entrie and add the new one
            S[0][j] = S[1][j];
            S[1][j] = Entrie;
 
            // Keep track of the max square length
            Max = max(Max, Entrie);
        }
     
    // Print the square
    cout << "Maximum size sub-matrix is: \n";
    for (int i = 0; i < Max; i++, cout << '\n')
        for (int j = 0; j < Max;j++)
            cout << "1 ";
}
 
// 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);
 
    return 0;
 
    // This code is contributed
    // by Gatea David
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG {
 
  static int R = 6 ;
  static int C = 5 ;
 
  static void printMaxSubSquare(int M[][])
  {
    int S[][] = new int[2][C], Max = 0;
 
    // set all elements of S to 0 first
    for (int i = 0; i < 2;i++){
      for (int j = 0; j < C;j++){
        S[i][j] =0;
      }
    }
 
    // Construct the entries
    for (int i = 0; i < R; i++){
      for (int j = 0; j < C; j++){
 
        // Compute the entrie at the current position
        int Entrie = M[i][j];
        if(Entrie != 0){
          if(j != 0){
            Entrie = 1 + Math.min(S[1][j - 1], Math.min(S[0][j - 1], S[1][j]));}}
 
        // Save the last entrie and add the new one
        S[0][j] = S[1][j];
        S[1][j] = Entrie;
 
        // Keep track of the max square length
        Max = Math.max(Max, Entrie);
      }
    }
 
    // Print the square
    System.out.print("Maximum size sub-matrix is: \n");
    for (int i = 0; i < Max; i++){
      for (int j = 0; j < Max;j++){
        System.out.print( "1 ");}
      System.out.println();
    }
  }
 
  // Driver Code
  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);
 
  }
}
 
// This code is contributed by code_hunt.

Javascript




<script>
 
// JavaScript code for Maximum size square
// sub-matrix with all 1s
// (space optimized solution)
const R = 6
const C = 5
 
function printMaxSubSquare(M)
{
    let Max = 0
    let S = new Array(2)
   
    // set all elements of S to 0 first
    for(let i=0;i<2;i++){
        S[i] = new Array(C).fill(0)
    }
 
    // Construct the entries
    for (let i = 0; i < R;i++)
        for (let j = 0; j < C;j++){
 
            // Compute the entrie at the current position
            let Entrie = M[i][j];
            if(Entrie)
                if(j)
                    Entrie = 1 + Math.min(S[1][j - 1],
                    Math.min(S[0][j - 1], S[1][j]));
 
            // Save the last entrie and add the new one
            S[0][j] = S[1][j];
            S[1][j] = Entrie;
 
            // Keep track of the max square length
            Max = Math.max(Max, Entrie);
        }
     
    // Print the square
    document.write("Maximum size sub-matrix is: ","</br>")
    for (let i = 0; i < Max; i++){
        for (let j = 0; j < Max;j++)
            document.write("1 ")
        document.write("</br>")
    }
         
}
 
// Driver code
const 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 shinjanpatra
 
</script>

Python3




# Python code for Maximum size square
# sub-matrix with all 1s
# (space optimized solution)
 
R = 6
C = 5
 
def printMaxSubSquare(M):
 
    global R,C
    Max = 0
   
    # set all elements of S to 0 first
    S = [[0 for col in range(C)]for row in range(2)]
 
    # Construct the entries
    for i in range(R):
        for j in range(C):
 
            # Compute the entrie at the current position
            Entrie = M[i][j]
            if(Entrie):
                if(j):
                    Entrie = 1 + min(S[1][j - 1],min(S[0][j - 1], S[1][j]))
 
            # Save the last entrie and add the new one
            S[0][j] = S[1][j]
            S[1][j] = Entrie
 
            # Keep track of the max square length
            Max = max(Max, Entrie)
     
    # Print the square
    print("Maximum size sub-matrix is: ")
    for i in range(Max):
        for j in range(Max):
            print("1",end=" ")
        print()
         
 
# Driver code
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 shinjanpatra

C#




// C# code to implement the approach
using System;
using System.Numerics;
using System.Collections.Generic;
 
public class GFG {
 
  static int R = 6 ;
  static int C = 5 ;
  
  static void printMaxSubSquare(int[,] M)
  {
    int[,] S = new int[2, C];
    int Maxx = 0;
  
    // set all elements of S to 0 first
    for (int i = 0; i < 2;i++){
      for (int j = 0; j < C;j++){
        S[i, j] =0;
      }
    }
  
    // Construct the entries
    for (int i = 0; i < R; i++){
      for (int j = 0; j < C; j++){
  
        // Compute the entrie at the current position
        int Entrie = M[i, j];
        if(Entrie != 0){
          if(j != 0){
            Entrie = 1 + Math.Min(S[1, j - 1], Math.Min(S[0, j - 1], S[1, j]));}}
  
        // Save the last entrie and add the new one
        S[0,j] = S[1,j];
        S[1,j] = Entrie;
  
        // Keep track of the max square length
        Maxx = Math.Max(Maxx, Entrie);
      }
    }
  
    // Print the square
    Console.Write("Maximum size sub-matrix is: \n");
    for (int i = 0; i < Maxx; i++){
      for (int j = 0; j < Maxx;j++){
        Console.Write( "1 ");}
      Console.WriteLine();
    }
  }
 
// Driver Code
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 the number of rows and n is the number of columns in the given matrix. Auxiliary space: O(n) where n is the number of columns in the given matrix. 

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


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