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 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:
" ; 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 << "
" ; } } /* 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:
" ); 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 ( "
" ); } } /* 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:
" ); 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 "
" ; } } # 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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