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++ 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 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 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 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# 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 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 ?> |
<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> |
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++ 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 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 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 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. |
# 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# 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);
}
} |
<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> |
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