Maximum size square sub-matrix with all 1s

Here given code implementation process.

// 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


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