Maximum size square sub-matrix with all 1s
The problem is to find the maximum size square sub-matrix with all ones in a given 2D matrix. A square sub-matrix is a contiguous sub-matrix where all the elements are ones.
Example
Consider the following 2D matrix:
0 1 1 1 1 0 1 1 0 1
1 1 1 0 0 0 1 1 1 1
1 1 1 1 1 0 1 1 0 1
1 1 1 1 0 0 1 0 0 1
0 1 1 1 1 1 1 1 0 1
0 0 0 1 1 1 1 0 0 1
1 1 0 1 1 1 1 1 0 1
0 1 1 1 1 1 1 1 0 1
1 1 1 1 0 0 0 1 0 1
0 1 1 1 1 0 1 1 0 1The maximum size square sub-matrix with all ones in this matrix is:
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1Idea to Solve the Problem
To find the maximum size square sub-matrix with all ones, we can use dynamic programming. We will create an auxiliary matrix with the same dimensions as the input matrix. Each cell in the auxiliary matrix will represent the size of the largest square sub-matrix with all ones that ends at that cell.
Algorithm
- Create an auxiliary matrix of the same dimensions as the input matrix, initialized with all zeroes.
- Traverse the first row and the first column of the input matrix. For each element in the first row or the first column that is equal to 1, update the corresponding cell in the auxiliary matrix to 1.
- For each cell (i, j) in the input matrix (excluding the first row and first column):
a. If the current cell (i, j) is equal to 1:
- Set the value of the corresponding cell in the auxiliary matrix (i, j) to the minimum of the values of the three adjacent cells (i, j-1), (i-1, j-1), and (i-1, j) in the auxiliary matrix, and add 1 to it. This represents the size of the largest square sub-matrix with all ones that ends at the current cell.
- Find the maximum value in the auxiliary matrix. The value of this cell (i, j) represents the size of the maximum square sub-matrix with all ones. The top-left corner of this sub-matrix will be at the location (i - (value - 1), j - (value - 1)).
Pseudocode
maximumSquare(matrix):
row = number of rows in matrix
col = number of columns in matrix
Create an auxiliary matrix of size row x col initialized with all zeroes
a = -1, b = -1
Traverse the first row and the first column of the matrix:
if the current cell is 1:
set the corresponding cell in the auxiliary matrix to 1
set a = i and b = j
for i from 1 to row-1:
for j from 1 to col-1:
if matrix[i][j] is 1:
auxiliary[i][j] = minimum(auxiliary[i][j-1], auxiliary[i-1][j-1], auxiliary[i-1][j]) + 1
if auxiliary[i][j] > auxiliary[a][b]:
set a = i and b = j
if a == -1 and b == -1:
print "None" (no square sub-matrix with all ones)
else:
print "Maximum Submatrix exists at location (" + (a - (auxiliary[a][b] - 1)) + "," + (b - (auxiliary[a][b] - 1)) + ") (" + a + "," + b + ")"
print "Maximum size : " + auxiliary[a][b]
Print the sub-matrix starting from (a - (auxiliary[a][b] - 1), b - (auxiliary[a][b] - 1)) to (a, b) in the input matrixCode Solution
// C Program
// Maximum size square sub-matrix with all 1s
#include <stdio.h>
// Matrix size
#define R 12
#define C 10
// Returns the minimum value of given two numbers (integer)
int minimum(int a, int b)
{
if (a > b)
{
return b;
}
else
{
return a;
}
}
// Finding the Maximum size rectangle binary sub-matrix with all 1s
void maximumSquare(int matrix[R][C])
{
// Auxiliary matrix which is used to find and store all square matrix
int auxiliary[R][C];
// Loop controllong variables
int i = 0;
int j = 0;
// This is used to detect square matrix location
int a = -1;
int b = -1;
// Set first element of each rows
for (i = 0; i < R; ++i)
{
auxiliary[i][0] = matrix[i][0];
}
// Set first element of each columns
for (j = 0; j < C; ++j)
{
auxiliary[0][j] = matrix[0][j];
}
for (i = 1; i < R; ++i)
{
for (j = 1; j < C; ++j)
{
// Set the initial element of matrix auxiliary [i][j] as zero
auxiliary[i][j] = 0;
if (matrix[i][j] == 1)
{
/*
Set x location value by minimum of side three elements (such as).
x [this is current location]
e2 e3
╲ │
╲ │
╲ │
e1‒‒‒ x
*/
auxiliary[i][j] = minimum(minimum(auxiliary[i][j - 1], auxiliary[i - 1][j - 1]), auxiliary[i - 1][j]) + 1;
if ((a == -1 && b == -1) || (auxiliary[i][j] > auxiliary[a][b]))
{
// When getting the new higher square matrix
a = i;
b = j;
}
}
}
}
if (a == -1 && b == -1)
{
printf("\n None ");
}
else
{
// Get top left location
i = a - (auxiliary[a][b] - 1);
j = b - (auxiliary[a][b] - 1);
// Display location of sub matrix
printf(" Maximum Submatrix exists at location (%d,%d) (%d,%d) ", i, j, a, b);
printf("\n Maximum size : %d \n", auxiliary[a][b]);
// Display result
// Rows
while (i <= a)
{
// Columns
while (j <= b)
{
printf(" %d", matrix[i][j]);
j++;
}
printf("\n");
// next row
i++;
// Get first resultant column
j = b - (auxiliary[a][b] - 1);
}
}
printf("\n");
}
int main(int argc, char
const *argv[])
{
// Define matrix of integer element which contains (0 and 1)
int matrix[R][C] =
{
{0, 1, 1, 1, 1, 0, 1, 1, 0, 1},
{1, 1, 1, 0, 0, 0, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 0, 1, 1, 0, 1},
{1, 1, 1, 1, 0, 0, 1, 0, 0, 1},
{0, 1, 1, 1, 1, 1, 1, 1, 0, 1},
{0, 0, 0, 1, 1, 1, 1, 0, 0, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 0, 1},
{0, 1, 1, 1, 1, 1, 1, 1, 0, 1},
{1, 1, 1, 1, 0, 0, 0, 1, 0, 1},
{0, 1, 1, 1, 1, 0, 1, 1, 0, 1}
};
maximumSquare(matrix);
return 0;
}Output
Maximum Submatrix exists at location (4,3) (7,6)
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
/*
Java Program for
Maximum size square sub-matrix with all 1s
*/
class SquareSubmatrix
{
// Returns the minimum value of given two numbers (integer)
public int minimum(int a, int b)
{
if (a > b)
{
return b;
}
else
{
return a;
}
}
// Finding the Maximum size rectangle binary sub-matrix with all 1s
public void maximumSquare(int[][] matrix)
{
int row = matrix.length;
int col = matrix[0].length;
// Auxiliary matrix which is used to find and store all square matrix
int[][] auxiliary = new int[row][col];
// Loop controllong variables
int i = 0;
int j = 0;
// This is used to detect square matrix location
int a = -1;
int b = -1;
// Set first element of each rows
for (i = 0; i < row; ++i)
{
auxiliary[i][0] = matrix[i][0];
}
// Set first element of each columns
for (j = 0; j < col; ++j)
{
auxiliary[0][j] = matrix[0][j];
}
for (i = 1; i < row; ++i)
{
for (j = 1; j < col; ++j)
{
// Set the initial element of matrix auxiliary [i][j] as zero
auxiliary[i][j] = 0;
if (matrix[i][j] == 1)
{
/*
Set x location value by minimum of side three elements (such as).
x [this is current location]
e2 e3
╲ │
╲ │
╲ │
e1‒‒‒ x
*/
auxiliary[i][j] = minimum(minimum(auxiliary[i][j - 1], auxiliary[i - 1][j - 1]), auxiliary[i - 1][j]) + 1;
if ((a == -1 && b == -1) || (auxiliary[i][j] > auxiliary[a][b]))
{
// When getting the new higher square matrix
a = i;
b = j;
}
}
}
}
if (a == -1 && b == -1)
{
System.out.print("\n None ");
}
else
{
// Get top left location
i = a - (auxiliary[a][b] - 1);
j = b - (auxiliary[a][b] - 1);
// Display location of sub matrix
System.out.print(" Maximum Submatrix exists at location (" + i + "," + j + ") (" + a + "," + b + ") ");
System.out.print("\n Maximum size : " + auxiliary[a][b] + " \n");
// Display result
// Rows
while (i <= a)
{
// Columns
while (j <= b)
{
System.out.print(" " + matrix[i][j] + "");
j++;
}
System.out.print("\n");
// next row
i++;
// Get first resultant column
j = b - (auxiliary[a][b] - 1);
}
}
System.out.print("\n");
}
public static void main(String[] args)
{
SquareSubmatrix task = new SquareSubmatrix();
// Define matrix of integer element which contains (0 and 1)
int[][] matrix =
{
{0, 1, 1, 1, 1, 0, 1, 1, 0, 1},
{1, 1, 1, 0, 0, 0, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 0, 1, 1, 0, 1},
{1, 1, 1, 1, 0, 0, 1, 0, 0, 1},
{0, 1, 1, 1, 1, 1, 1, 1, 0, 1},
{0, 0, 0, 1, 1, 1, 1, 0, 0, 1},
{1, 1, 0, 1, 1, 1, 1, 1, 0, 1},
{0, 1, 1, 1, 1, 1, 1, 1, 0, 1},
{1, 1, 1, 1, 0, 0, 0, 1, 0, 1},
{0, 1, 1, 1, 1, 0, 1, 1, 0, 1}
};
task.maximumSquare(matrix);
}
}Output
Maximum Submatrix exists at location (4,3) (7,6)
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
// Include header file
#include <iostream>
// Matrix size
#define R 12
#define C 10
using namespace std;
/*
C++ Program for
Maximum size square sub-matrix with all 1s
*/
class SquareSubmatrix
{
public:
// Returns the minimum value of given two numbers (integer)
int minimum(int a, int b)
{
if (a > b)
{
return b;
}
else
{
return a;
}
}
// Finding the Maximum size rectangle binary sub-matrix with all 1s
void maximumSquare(int matrix[R][C])
{
// Auxiliary matrix which is used to find and store all square matrix
int auxiliary[R][C];
// Loop controllong variables
int i = 0;
int j = 0;
// This is used to detect square matrix location
int a = -1;
int b = -1;
// Set first element of each rows
for (i = 0; i < R; ++i)
{
auxiliary[i][0] = matrix[i][0];
}
// Set first element of each columns
for (j = 0; j < C; ++j)
{
auxiliary[0][j] = matrix[0][j];
}
for (i = 1; i < R; ++i)
{
for (j = 1; j < C; ++j)
{
// Set the initial element of matrix auxiliary [i][j] as zero
auxiliary[i][j] = 0;
if (matrix[i][j] == 1)
{
/*
Set x location value by minimum of side three elements (such as).
x [this is current location]
e2 e3
╲ │
╲ │
╲ │
e1‒‒‒ x
*/
auxiliary[i][j] = this->minimum(this->minimum(auxiliary[i][j - 1], auxiliary[i - 1][j - 1]), auxiliary[i - 1][j]) + 1;
if ((a == -1 && b == -1) || (auxiliary[i][j] > auxiliary[a][b]))
{
// When getting the new higher square matrix
a = i;
b = j;
}
}
}
}
if (a == -1 && b == -1)
{
cout << "\n None ";
}
else
{
// Get top left location
i = a - (auxiliary[a][b] - 1);
j = b - (auxiliary[a][b] - 1);
// Display location of sub matrix
cout << " Maximum Submatrix exists at location (" << i << "," << j << ") (" << a << "," << b << ") ";
cout << "\n Maximum size : " << auxiliary[a][b] << " \n";
// Display result
// Rows
while (i <= a)
{
// Columns
while (j <= b)
{
cout << " " << matrix[i][j] << "";
j++;
}
cout << "\n";
// next row
i++;
// Get first resultant column
j = b - (auxiliary[a][b] - 1);
}
}
cout << "\n";
}
};
int main()
{
SquareSubmatrix task = SquareSubmatrix();
// Define matrix of integer element which contains (0 and 1)
int matrix[R][C] =
{
{0, 1, 1, 1, 1, 0, 1, 1, 0, 1},
{1, 1, 1, 0, 0, 0, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 0, 1, 1, 0, 1},
{1, 1, 1, 1, 0, 0, 1, 0, 0, 1},
{0, 1, 1, 1, 1, 1, 1, 1, 0, 1},
{0, 0, 0, 1, 1, 1, 1, 0, 0, 1},
{1, 1, 0, 1, 1, 1, 1, 1, 0, 1},
{0, 1, 1, 1, 1, 1, 1, 1, 0, 1},
{1, 1, 1, 1, 0, 0, 0, 1, 0, 1},
{0, 1, 1, 1, 1, 0, 1, 1, 0, 1}
};
task.maximumSquare(matrix);
return 0;
}Output
Maximum Submatrix exists at location (4,3) (7,6)
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
// Include namespace system
using System;
/*
C# Program for
Maximum size square sub-matrix with all 1s
*/
public class SquareSubmatrix
{
// Returns the minimum value of given two numbers (integer)
public int minimum(int a, int b)
{
if (a > b)
{
return b;
}
else
{
return a;
}
}
// Finding the Maximum size rectangle binary sub-matrix with all 1s
public void maximumSquare(int[,] matrix)
{
int row = matrix.GetLength(0);
int col = matrix.GetLength(1);
// Auxiliary matrix which is used to find and store all square matrix
int[,] auxiliary = new int[row,col];
// Loop controllong variables
int i = 0;
int j = 0;
// This is used to detect square matrix location
int a = -1;
int b = -1;
// Set first element of each rows
for (i = 0; i < row; ++i)
{
auxiliary[i,0] = matrix[i,0];
}
// Set first element of each columns
for (j = 0; j < col; ++j)
{
auxiliary[0,j] = matrix[0,j];
}
for (i = 1; i < row; ++i)
{
for (j = 1; j < col; ++j)
{
// Set the initial element of matrix auxiliary [i,j] as zero
auxiliary[i,j] = 0;
if (matrix[i,j] == 1)
{
/*
Set x location value by minimum of side three elements (such as).
x [this is current location]
e2 e3
╲ │
╲ │
╲ │
e1‒‒‒ x
*/
auxiliary[i,j] = minimum(minimum(auxiliary[i,j - 1], auxiliary[i - 1,j - 1]), auxiliary[i - 1,j]) + 1;
if ((a == -1 && b == -1) || (auxiliary[i,j] > auxiliary[a,b]))
{
// When getting the new higher square matrix
a = i;
b = j;
}
}
}
}
if (a == -1 && b == -1)
{
Console.Write("\n None ");
}
else
{
// Get top left location
i = a - (auxiliary[a,b] - 1);
j = b - (auxiliary[a,b] - 1);
// Display location of sub matrix
Console.Write(" Maximum Submatrix exists at location (" + i + "," + j + ") (" + a + "," + b + ") ");
Console.Write("\n Maximum size : " + auxiliary[a,b] + " \n");
// Display result
// Rows
while (i <= a)
{
// Columns
while (j <= b)
{
Console.Write(" " + matrix[i,j] + "");
j++;
}
Console.Write("\n");
// next row
i++;
// Get first resultant column
j = b - (auxiliary[a,b] - 1);
}
}
Console.Write("\n");
}
public static void Main(String[] args)
{
SquareSubmatrix task = new SquareSubmatrix();
// Define matrix of integer element which contains (0 and 1)
int[,] matrix = {
{
0 , 1 , 1 , 1 , 1 , 0 , 1 , 1 , 0 , 1
} , {
1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 1
} , {
1 , 1 , 1 , 1 , 1 , 0 , 1 , 1 , 0 , 1
} , {
1 , 1 , 1 , 1 , 0 , 0 , 1 , 0 , 0 , 1
} , {
0 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 0 , 1
} , {
0 , 0 , 0 , 1 , 1 , 1 , 1 , 0 , 0 , 1
} , {
1 , 1 , 0 , 1 , 1 , 1 , 1 , 1 , 0 , 1
} , {
0 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 0 , 1
} , {
1 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 0 , 1
} , {
0 , 1 , 1 , 1 , 1 , 0 , 1 , 1 , 0 , 1
}
};
task.maximumSquare(matrix);
}
}Output
Maximum Submatrix exists at location (4,3) (7,6)
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
<?php
/*
Php Program for
Maximum size square sub-matrix with all 1s
*/
class SquareSubmatrix
{
// Returns the minimum value of given two numbers (integer)
public function minimum($a, $b)
{
if ($a > $b)
{
return $b;
}
else
{
return $a;
}
}
// Finding the Maximum size rectangle binary sub-matrix with all 1s
public function maximumSquare( $matrix)
{
$row = count($matrix);
$col = count($matrix[0]);
// Auxiliary matrix which is used to find and store all square matrix
$auxiliary = array_fill(0, $row, array_fill(0, $col, 0));
// Loop controllong variables
$i = 0;
$j = 0;
// This is used to detect square matrix location
$a = -1;
$b = -1;
// Set first element of each rows
for ($i = 0; $i < $row; ++$i)
{
$auxiliary[$i][0] = $matrix[$i][0];
}
// Set first element of each columns
for ($j = 0; $j < $col; ++$j)
{
$auxiliary[0][$j] = $matrix[0][$j];
}
for ($i = 1; $i < $row; ++$i)
{
for ($j = 1; $j < $col; ++$j)
{
// Set the initial element of matrix auxiliary [i][j] as zero
$auxiliary[$i][$j] = 0;
if ($matrix[$i][$j] == 1)
{
/*
Set x location value by minimum of side three elements (such as).
x [this is current location]
e2 e3
╲ │
╲ │
╲ │
e1‒‒‒ x
*/
$auxiliary[$i][$j] = $this->minimum($this->minimum($auxiliary[$i][$j - 1], $auxiliary[$i - 1][$j - 1]), $auxiliary[$i - 1][$j]) + 1;
if (($a == -1 && $b == -1) || ($auxiliary[$i][$j] > $auxiliary[$a][$b]))
{
// When getting the new higher square matrix
$a = $i;
$b = $j;
}
}
}
}
if ($a == -1 && $b == -1)
{
echo "\n None ";
}
else
{
// Get top left location
$i = $a - ($auxiliary[$a][$b] - 1);
$j = $b - ($auxiliary[$a][$b] - 1);
// Display location of sub matrix
echo " Maximum Submatrix exists at location (". $i .",". $j .") (". $a .",". $b .") ";
echo "\n Maximum size : ". $auxiliary[$a][$b] ." \n";
// Display result
// Rows
while ($i <= $a)
{
// Columns
while ($j <= $b)
{
echo " ". $matrix[$i][$j] ."";
$j++;
}
echo "\n";
// next row
$i++;
// Get first resultant column
$j = $b - ($auxiliary[$a][$b] - 1);
}
}
echo "\n";
}
}
function main()
{
$task = new SquareSubmatrix();
// Define matrix of integer element which contains (0 and 1)
$matrix = array(
array(0, 1, 1, 1, 1, 0, 1, 1, 0, 1),
array(1, 1, 1, 0, 0, 0, 1, 1, 1, 1),
array(1, 1, 1, 1, 1, 0, 1, 1, 0, 1),
array(1, 1, 1, 1, 0, 0, 1, 0, 0, 1),
array(0, 1, 1, 1, 1, 1, 1, 1, 0, 1),
array(0, 0, 0, 1, 1, 1, 1, 0, 0, 1),
array(1, 1, 0, 1, 1, 1, 1, 1, 0, 1),
array(0, 1, 1, 1, 1, 1, 1, 1, 0, 1),
array(1, 1, 1, 1, 0, 0, 0, 1, 0, 1),
array(0, 1, 1, 1, 1, 0, 1, 1, 0, 1)
);
$task->maximumSquare($matrix);
}
main();Output
Maximum Submatrix exists at location (4,3) (7,6)
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
/*
Node Js Program for
Maximum size square sub-matrix with all 1s
*/
class SquareSubmatrix
{
// Returns the minimum value of given two numbers (integer)
minimum(a, b)
{
if (a > b)
{
return b;
}
else
{
return a;
}
}
// Finding the Maximum size rectangle binary sub-matrix with all 1s
maximumSquare(matrix)
{
var row = matrix.length;
var col = matrix[0].length;
// Auxiliary matrix which is used to find and store all square matrix
var auxiliary = Array(row).fill(0).map(() => new Array(col).fill(0));
// Loop controllong variables
var i = 0;
var j = 0;
// This is used to detect square matrix location
var a = -1;
var b = -1;
// Set first element of each rows
for (i = 0; i < row; ++i)
{
auxiliary[i][0] = matrix[i][0];
}
// Set first element of each columns
for (j = 0; j < col; ++j)
{
auxiliary[0][j] = matrix[0][j];
}
for (i = 1; i < row; ++i)
{
for (j = 1; j < col; ++j)
{
// Set the initial element of matrix auxiliary [i][j] as zero
auxiliary[i][j] = 0;
if (matrix[i][j] == 1)
{
/*
Set x location value by minimum of side three elements (such as).
x [this is current location]
e2 e3
╲ │
╲ │
╲ │
e1‒‒‒ x
*/
auxiliary[i][j] = this.minimum(this.minimum(auxiliary[i][j - 1], auxiliary[i - 1][j - 1]), auxiliary[i - 1][j]) + 1;
if ((a == -1 && b == -1) || (auxiliary[i][j] > auxiliary[a][b]))
{
// When getting the new higher square matrix
a = i;
b = j;
}
}
}
}
if (a == -1 && b == -1)
{
process.stdout.write("\n None ");
}
else
{
// Get top left location
i = a - (auxiliary[a][b] - 1);
j = b - (auxiliary[a][b] - 1);
// Display location of sub matrix
process.stdout.write(" Maximum Submatrix exists at location (" + i + "," + j + ") (" + a + "," + b + ") ");
process.stdout.write("\n Maximum size : " + auxiliary[a][b] + " \n");
// Display result
// Rows
while (i <= a)
{
// Columns
while (j <= b)
{
process.stdout.write(" " + matrix[i][j] + "");
j++;
}
process.stdout.write("\n");
// next row
i++;
// Get first resultant column
j = b - (auxiliary[a][b] - 1);
}
}
process.stdout.write("\n");
}
}
function main()
{
var task = new SquareSubmatrix();
// Define matrix of integer element which contains (0 and 1)
var matrix = [
[0, 1, 1, 1, 1, 0, 1, 1, 0, 1] ,
[1, 1, 1, 0, 0, 0, 1, 1, 1, 1] ,
[1, 1, 1, 1, 1, 0, 1, 1, 0, 1] ,
[1, 1, 1, 1, 0, 0, 1, 0, 0, 1] ,
[0, 1, 1, 1, 1, 1, 1, 1, 0, 1] ,
[0, 0, 0, 1, 1, 1, 1, 0, 0, 1] ,
[1, 1, 0, 1, 1, 1, 1, 1, 0, 1] ,
[0, 1, 1, 1, 1, 1, 1, 1, 0, 1] ,
[1, 1, 1, 1, 0, 0, 0, 1, 0, 1] ,
[0, 1, 1, 1, 1, 0, 1, 1, 0, 1]
];
task.maximumSquare(matrix);
}
main();Output
Maximum Submatrix exists at location (4,3) (7,6)
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
# Python 3 Program for
# Maximum size square sub-matrix with all 1s
class SquareSubmatrix :
# Returns the minimum value of given two numbers (integer)
def minimum(self, a, b) :
if (a > b) :
return b
else :
return a
# Finding the Maximum size rectangle binary sub-matrix with all 1s
def maximumSquare(self, matrix) :
row = len(matrix)
col = len(matrix[0])
# Auxiliary matrix which is used to find and store all square matrix
auxiliary = [[0] * (col) for _ in range(row) ]
# Loop controllong variables
i = 0
j = 0
# This is used to detect square matrix location
a = -1
b = -1
# Set first element of each rows
while (i < row) :
auxiliary[i][0] = matrix[i][0]
i += 1
# Set first element of each columns
while (j < col) :
auxiliary[0][j] = matrix[0][j]
j += 1
i = 1
while (i < row) :
j = 1
while (j < col) :
# Set the initial element of matrix auxiliary [i][j] as zero
auxiliary[i][j] = 0
if (matrix[i][j] == 1) :
#
# Set x location value by minimum of side three elements (such as).
# x [this is current location]
# e2 e3
# ╲ │
# ╲ │
# ╲ │
# e1‒‒‒ x
auxiliary[i][j] = self.minimum(self.minimum(auxiliary[i][j - 1], auxiliary[i - 1][j - 1]), auxiliary[i - 1][j]) + 1
if ((a == -1 and b == -1) or(auxiliary[i][j] > auxiliary[a][b])) :
# When getting the new higher square matrix
a = i
b = j
j += 1
i += 1
if (a == -1 and b == -1) :
print("\n None ", end = "")
else :
# Get top left location
i = a - (auxiliary[a][b] - 1)
j = b - (auxiliary[a][b] - 1)
# Display location of sub matrix
print(" Maximum Submatrix exists at location (", i ,",", j ,") (", a ,",", b ,") ", end = "")
print("\n Maximum size : ", auxiliary[a][b] ," ")
# Display result
# Rows
while (i <= a) :
# Columns
while (j <= b) :
print(" ", matrix[i][j] ,"", end = "")
j += 1
print(end = "\n")
# next row
i += 1
# Get first resultant column
j = b - (auxiliary[a][b] - 1)
print(end = "\n")
def main() :
task = SquareSubmatrix()
# Define matrix of integer element which contains (0 and 1)
matrix = [
[0, 1, 1, 1, 1, 0, 1, 1, 0, 1] ,
[1, 1, 1, 0, 0, 0, 1, 1, 1, 1] ,
[1, 1, 1, 1, 1, 0, 1, 1, 0, 1] ,
[1, 1, 1, 1, 0, 0, 1, 0, 0, 1] ,
[0, 1, 1, 1, 1, 1, 1, 1, 0, 1] ,
[0, 0, 0, 1, 1, 1, 1, 0, 0, 1] ,
[1, 1, 0, 1, 1, 1, 1, 1, 0, 1] ,
[0, 1, 1, 1, 1, 1, 1, 1, 0, 1] ,
[1, 1, 1, 1, 0, 0, 0, 1, 0, 1] ,
[0, 1, 1, 1, 1, 0, 1, 1, 0, 1]
]
task.maximumSquare(matrix)
if __name__ == "__main__": main()Output
Maximum Submatrix exists at location ( 4 , 3 ) ( 7 , 6 )
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
# Ruby Program for
# Maximum size square sub-matrix with all 1s
class SquareSubmatrix
# Returns the minimum value of given two numbers (integer)
def minimum(a, b)
if (a > b)
return b
else
return a
end
end
# Finding the Maximum size rectangle binary sub-matrix with all 1s
def maximumSquare(matrix)
row = matrix.length
col = matrix[0].length
# Auxiliary matrix which is used to find and store all square matrix
auxiliary = Array.new(row) {Array.new(col) {0}}
# Loop controllong variables
i = 0
j = 0
# This is used to detect square matrix location
a = -1
b = -1
# Set first element of each rows
while (i < row)
auxiliary[i][0] = matrix[i][0]
i += 1
end
# Set first element of each columns
while (j < col)
auxiliary[0][j] = matrix[0][j]
j += 1
end
i = 1
while (i < row)
j = 1
while (j < col)
# Set the initial element of matrix auxiliary [i][j] as zero
auxiliary[i][j] = 0
if (matrix[i][j] == 1)
#
# Set x location value by minimum of side three elements (such as).
# x [this is current location]
# e2 e3
# ╲ │
# ╲ │
# ╲ │
# e1‒‒‒ x
auxiliary[i][j] = self.minimum(self.minimum(auxiliary[i][j - 1], auxiliary[i - 1][j - 1]), auxiliary[i - 1][j]) + 1
if ((a == -1 && b == -1) || (auxiliary[i][j] > auxiliary[a][b]))
# When getting the new higher square matrix
a = i
b = j
end
end
j += 1
end
i += 1
end
if (a == -1 && b == -1)
print("\n None ")
else
# Get top left location
i = a - (auxiliary[a][b] - 1)
j = b - (auxiliary[a][b] - 1)
# Display location of sub matrix
print(" Maximum Submatrix exists at location (", i ,",", j ,") (", a ,",", b ,") ")
print("\n Maximum size : ", auxiliary[a][b] ," \n")
# Display result
# Rows
while (i <= a)
# Columns
while (j <= b)
print(" ", matrix[i][j] ,"")
j += 1
end
print("\n")
# next row
i += 1
# Get first resultant column
j = b - (auxiliary[a][b] - 1)
end
end
print("\n")
end
end
def main()
task = SquareSubmatrix.new()
# Define matrix of integer element which contains (0 and 1)
matrix = [
[0, 1, 1, 1, 1, 0, 1, 1, 0, 1] ,
[1, 1, 1, 0, 0, 0, 1, 1, 1, 1] ,
[1, 1, 1, 1, 1, 0, 1, 1, 0, 1] ,
[1, 1, 1, 1, 0, 0, 1, 0, 0, 1] ,
[0, 1, 1, 1, 1, 1, 1, 1, 0, 1] ,
[0, 0, 0, 1, 1, 1, 1, 0, 0, 1] ,
[1, 1, 0, 1, 1, 1, 1, 1, 0, 1] ,
[0, 1, 1, 1, 1, 1, 1, 1, 0, 1] ,
[1, 1, 1, 1, 0, 0, 0, 1, 0, 1] ,
[0, 1, 1, 1, 1, 0, 1, 1, 0, 1]
]
task.maximumSquare(matrix)
end
main()Output
Maximum Submatrix exists at location (4,3) (7,6)
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
/*
Scala Program for
Maximum size square sub-matrix with all 1s
*/
class SquareSubmatrix
{
// Returns the minimum value of given two numbers (integer)
def minimum(a: Int, b: Int): Int = {
if (a > b)
{
return b;
}
else
{
return a;
}
}
// Finding the Maximum size rectangle binary sub-matrix with all 1s
def maximumSquare(matrix: Array[Array[Int]]): Unit = {
var row: Int = matrix.length;
var col: Int = matrix(0).length;
// Auxiliary matrix which is used to find and store all square matrix
var auxiliary: Array[Array[Int]] = Array.fill[Int](row, col)(0);
// Loop controllong variables
var i: Int = 0;
var j: Int = 0;
// This is used to detect square matrix location
var a: Int = -1;
var b: Int = -1;
// Set first element of each rows
while (i < row)
{
auxiliary(i)(0) = matrix(i)(0);
i += 1;
}
// Set first element of each columns
while (j < col)
{
auxiliary(0)(j) = matrix(0)(j);
j += 1;
}
i = 1;
while (i < row)
{
j = 1;
while (j < col)
{
// Set the initial element of matrix auxiliary [i][j] as zero
auxiliary(i)(j) = 0;
if (matrix(i)(j) == 1)
{
/*
Set x location value by minimum of side three elements (such as).
x [this is current location]
e2 e3
╲ │
╲ │
╲ │
e1‒‒‒ x
*/
auxiliary(i)(j) = this.minimum(this.minimum(auxiliary(i)(j - 1), auxiliary(i - 1)(j - 1)), auxiliary(i - 1)(j)) + 1;
if ((a == -1 && b == -1) || (auxiliary(i)(j) > auxiliary(a)(b)))
{
// When getting the new higher square matrix
a = i;
b = j;
}
}
j += 1;
}
i += 1;
}
if (a == -1 && b == -1)
{
print("\n None ");
}
else
{
// Get top left location
i = a - (auxiliary(a)(b) - 1);
j = b - (auxiliary(a)(b) - 1);
// Display location of sub matrix
print(" Maximum Submatrix exists at location (" + i + "," + j + ") (" + a + "," + b + ") ");
print("\n Maximum size : " + auxiliary(a)(b) + " \n");
// Display result
// Rows
while (i <= a)
{
// Columns
while (j <= b)
{
print(" " + matrix(i)(j) + "");
j += 1;
}
print("\n");
// next row
i += 1;
// Get first resultant column
j = b - (auxiliary(a)(b) - 1);
}
}
print("\n");
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: SquareSubmatrix = new SquareSubmatrix();
// Define matrix of integer element which contains (0 and 1)
var matrix: Array[Array[Int]] = Array(
Array(0, 1, 1, 1, 1, 0, 1, 1, 0, 1),
Array(1, 1, 1, 0, 0, 0, 1, 1, 1, 1),
Array(1, 1, 1, 1, 1, 0, 1, 1, 0, 1),
Array(1, 1, 1, 1, 0, 0, 1, 0, 0, 1),
Array(0, 1, 1, 1, 1, 1, 1, 1, 0, 1),
Array(0, 0, 0, 1, 1, 1, 1, 0, 0, 1),
Array(1, 1, 0, 1, 1, 1, 1, 1, 0, 1),
Array(0, 1, 1, 1, 1, 1, 1, 1, 0, 1),
Array(1, 1, 1, 1, 0, 0, 0, 1, 0, 1),
Array(0, 1, 1, 1, 1, 0, 1, 1, 0, 1)
);
task.maximumSquare(matrix);
}
}Output
Maximum Submatrix exists at location (4,3) (7,6)
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
/*
Swift 4 Program for
Maximum size square sub-matrix with all 1s
*/
class SquareSubmatrix
{
// Returns the minimum value of given two numbers (integer)
func minimum(_ a: Int, _ b: Int)->Int
{
if (a > b)
{
return b;
}
else
{
return a;
}
}
// Finding the Maximum size rectangle binary sub-matrix with all 1s
func maximumSquare(_ matrix: [[Int]])
{
let row: Int = matrix.count;
let col: Int = matrix[0].count;
// Auxiliary matrix which is used to find and store all square matrix
var auxiliary: [[Int]] = Array(repeating: Array(repeating: 0, count: col), count: row);
// Loop controllong variables
var i: Int = 0;
var j: Int = 0;
// This is used to detect square matrix location
var a: Int = -1;
var b: Int = -1;
// Set first element of each rows
while (i < row)
{
auxiliary[i][0] = matrix[i][0];
i += 1;
}
// Set first element of each columns
while (j < col)
{
auxiliary[0][j] = matrix[0][j];
j += 1;
}
i = 1;
while (i < row)
{
j = 1;
while (j < col)
{
// Set the initial element of matrix auxiliary [i][j] as zero
auxiliary[i][j] = 0;
if (matrix[i][j] == 1)
{
/*
Set x location value by minimum of side three elements (such as).
x [this is current location]
e2 e3
╲ │
╲ │
╲ │
e1‒‒‒ x
*/
auxiliary[i][j] = self.minimum(self.minimum(auxiliary[i][j - 1], auxiliary[i - 1][j - 1]), auxiliary[i - 1][j]) + 1;
if ((a == -1 && b == -1) || (auxiliary[i][j] > auxiliary[a][b]))
{
// When getting the new higher square matrix
a = i;
b = j;
}
}
j += 1;
}
i += 1;
}
if (a == -1 && b == -1)
{
print("\n None ", terminator: "");
}
else
{
// Get top left location
i = a - (auxiliary[a][b] - 1);
j = b - (auxiliary[a][b] - 1);
// Display location of sub matrix
print(" Maximum Submatrix exists at location (", i ,",", j ,") (", a ,",", b ,") ", terminator: "");
print("\n Maximum size : ", auxiliary[a][b] ," ");
// Display result
// Rows
while (i <= a)
{
// Columns
while (j <= b)
{
print(" ", matrix[i][j] ,"", terminator: "");
j += 1;
}
print(terminator: "\n");
// next row
i += 1;
// Get first resultant column
j = b - (auxiliary[a][b] - 1);
}
}
print(terminator: "\n");
}
}
func main()
{
let task: SquareSubmatrix = SquareSubmatrix();
// Define matrix of integer element which contains (0 and 1)
let matrix: [[Int]] = [
[0, 1, 1, 1, 1, 0, 1, 1, 0, 1] ,
[1, 1, 1, 0, 0, 0, 1, 1, 1, 1] ,
[1, 1, 1, 1, 1, 0, 1, 1, 0, 1] ,
[1, 1, 1, 1, 0, 0, 1, 0, 0, 1] ,
[0, 1, 1, 1, 1, 1, 1, 1, 0, 1] ,
[0, 0, 0, 1, 1, 1, 1, 0, 0, 1] ,
[1, 1, 0, 1, 1, 1, 1, 1, 0, 1] ,
[0, 1, 1, 1, 1, 1, 1, 1, 0, 1] ,
[1, 1, 1, 1, 0, 0, 0, 1, 0, 1] ,
[0, 1, 1, 1, 1, 0, 1, 1, 0, 1]
];
task.maximumSquare(matrix);
}
main();Output
Maximum Submatrix exists at location ( 4 , 3 ) ( 7 , 6 )
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
/*
Kotlin Program for
Maximum size square sub-matrix with all 1s
*/
class SquareSubmatrix
{
// Returns the minimum value of given two numbers (integer)
fun minimum(a: Int, b: Int): Int
{
if (a > b)
{
return b;
}
else
{
return a;
}
}
// Finding the Maximum size rectangle binary sub-matrix with all 1s
fun maximumSquare(matrix: Array<Array<Int>> ): Unit
{
var row: Int = matrix.count();
var col: Int = matrix[0].count();
// Auxiliary matrix which is used to find and store all square matrix
var auxiliary: Array < Array < Int >> = Array(row)
{
Array(col)
{
0
}
};
// Loop controllong variables
var i: Int = 0;
var j: Int = 0;
// This is used to detect square matrix location
var a: Int = -1;
var b: Int = -1;
// Set first element of each rows
while (i < row)
{
auxiliary[i][0] = matrix[i][0];
i += 1;
}
// Set first element of each columns
while (j < col)
{
auxiliary[0][j] = matrix[0][j];
j += 1;
}
i = 1;
while (i < row)
{
j = 1;
while (j < col)
{
// Set the initial element of matrix auxiliary [i][j] as zero
auxiliary[i][j] = 0;
if (matrix[i][j] == 1)
{
/*
Set x location value by minimum of side three elements (such as).
x [this is current location]
e2 e3
╲ │
╲ │
╲ │
e1‒‒‒ x
*/
auxiliary[i][j] = this.minimum(this.minimum(auxiliary[i][j - 1], auxiliary[i - 1][j - 1]), auxiliary[i - 1][j]) + 1;
if ((a == -1 && b == -1) || (auxiliary[i][j] > auxiliary[a][b]))
{
// When getting the new higher square matrix
a = i;
b = j;
}
}
j += 1;
}
i += 1;
}
if (a == -1 && b == -1)
{
print("\n None ");
}
else
{
// Get top left location
i = a - (auxiliary[a][b] - 1);
j = b - (auxiliary[a][b] - 1);
// Display location of sub matrix
print(" Maximum Submatrix exists at location (" + i + "," + j + ") (" + a + "," + b + ") ");
print("\n Maximum size : " + auxiliary[a][b] + " \n");
// Display result
// Rows
while (i <= a)
{
// Columns
while (j <= b)
{
print(" " + matrix[i][j] + "");
j += 1;
}
print("\n");
// next row
i += 1;
// Get first resultant column
j = b - (auxiliary[a][b] - 1);
}
}
print("\n");
}
}
fun main(args: Array < String > ): Unit
{
var task: SquareSubmatrix = SquareSubmatrix();
// Define matrix of integer element which contains (0 and 1)
var matrix: Array<Array<Int>> = arrayOf(
arrayOf(0, 1, 1, 1, 1, 0, 1, 1, 0, 1),
arrayOf(1, 1, 1, 0, 0, 0, 1, 1, 1, 1),
arrayOf(1, 1, 1, 1, 1, 0, 1, 1, 0, 1),
arrayOf(1, 1, 1, 1, 0, 0, 1, 0, 0, 1),
arrayOf(0, 1, 1, 1, 1, 1, 1, 1, 0, 1),
arrayOf(0, 0, 0, 1, 1, 1, 1, 0, 0, 1),
arrayOf(1, 1, 0, 1, 1, 1, 1, 1, 0, 1),
arrayOf(0, 1, 1, 1, 1, 1, 1, 1, 0, 1),
arrayOf(1, 1, 1, 1, 0, 0, 0, 1, 0, 1),
arrayOf(0, 1, 1, 1, 1, 0, 1, 1, 0, 1)
);
task.maximumSquare(matrix);
}Output
Maximum Submatrix exists at location (4,3) (7,6)
Maximum size : 4
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
Output Explanation
The java code implements the above algorithm to find the maximum size square sub-matrix with all ones in the given 2D matrix.
The output shows the location of the maximum square sub-matrix with all ones and its size. It also displays the sub-matrix.
Time Complexity
The time complexity of the provided solution is O(row * col), where 'row' is the number of rows and 'col' is the number of columns in the 2D matrix. The function traverses each element in the matrix once, and each operation takes constant time. Therefore, the overall time complexity is linear in the size of the matrix.
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