Given a binary matrix, find out the maximum size square sub-matrix with all 1s.
For example, consider the below binary matrix.
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 5using 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 5void 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 1spublic 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 1sdef 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 ProgramM = [[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 1susing 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 5void 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 entry 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 entry 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 codeint 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 approachimport 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. |
Python3
# Python code for Maximum size square# sub-matrix with all 1s# (space optimized solution)R = 6C = 5def 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 codeM = [[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 approachusing 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); }} |
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 codeconst 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> |
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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