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Check if matrix A can be converted to B by changing parity of corner elements of any submatrix

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  • Last Updated : 10 Dec, 2021
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Given two binary matrices, A[][] and B[][] of N×M. In a single operation, one can choose a sub-matrix (min of 2 rows and 2c columns) and change the parity of the corner elements i.e. 1 can be changed to a 0, and 0 can be changed to a 1. The task is to check if the matrix A can be converted to B using any number of operations.

Examples:  

Input: A[][] = {{0, 1, 0}, {0, 1, 0}, {1, 0, 0}}, 
B[][] = {{1, 0, 0}, {1, 0, 0}, {1, 0, 0}} 
Output: Yes 
Choose the sub-matrix whose top-left corner is (0, 0) 
and bottom-right corner is (1, 1). 
Changing the parity of the corner elements 
of this sub-matrix will convert A[][] to B[][]
Input: A[][] = {{0, 1, 0, 1}, {1, 0, 1, 0}, {0, 1, 0, 1}}, 
B[][] = {{1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}} 
Output: No 

Approach: The first thing to notice is that despite the conversions the total parity of both the matrix will remain the same. Hence, check if a[i][j] is not same as b[i][j] then change the parity of the corner elements of the sub-matrix whose left top corner is (0, 0) and right bottom corner is (i, j) for 1 ≤ i, j. After performing parity change for every a[i][j] which is not equal to b[i][j], check if both the matrix are the same or not. If they are the same, we can change A to B. 
Below is the implementation of the above approach: 

C++




// C++ implementation of the
// above approach
 
#include <bits/stdc++.h>
using namespace std;
#define N 3
#define M 3
 
// Boolean function that returns
// true or false
bool check(int a[N][M], int b[N][M])
{
    // Traverse for all elements
    for (int i = 1; i < N; i++) {
        for (int j = 1; j < M; j++) {
 
            // If both are not equal
            if (a[i][j] != b[i][j]) {
 
                // Change the parity of
                // all corner elements
                a[i][j] ^= 1;
                a[0][0] ^= 1;
                a[0][j] ^= 1;
                a[i][0] ^= 1;
            }
        }
    }
 
    // Check if A is equal to B
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
 
            // Not equal
            if (a[i][j] != b[i][j])
                return false;
        }
    }
    return true;
}
 
// Driver Code
int main()
{
    // First binary matrix
    int a[N][N] = { { 0, 1, 0 },
                    { 0, 1, 0 },
                    { 1, 0, 0 } };
 
    // Second binary matrix
    int b[N][N] = { { 1, 0, 0 },
                    { 1, 0, 0 },
                    { 1, 0, 0 } };
 
    if (check(a, b))
        cout << "Yes";
    else
        cout << "No";
}

Java




// Java implementation of the
// above approach
class GFG
{
     
static final int N = 3,M =3;
 
// Boolean function that returns
// true or false
static boolean check(int a[][], int b[][])
{
    // Traverse for all elements
    for (int i = 1; i < N; i++)
    {
        for (int j = 1; j < M; j++)
        {
 
            // If both are not equal
            if (a[i][j] != b[i][j])
            {
 
                // Change the parity of
                // all corner elements
                a[i][j] ^= 1;
                a[0][0] ^= 1;
                a[0][j] ^= 1;
                a[i][0] ^= 1;
            }
        }
    }
 
    // Check if A is equal to B
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
 
            // Not equal
            if (a[i][j] != b[i][j])
                return false;
        }
    }
    return true;
}
 
// Driver Code
public static void main(String args[])
{
    // First binary matrix
    int a[][] = { { 0, 1, 0 },
                    { 0, 1, 0 },
                    { 1, 0, 0 } };
 
    // Second binary matrix
    int b[][] = { { 1, 0, 0 },
                    { 1, 0, 0 },
                    { 1, 0, 0 } };
 
    if (check(a, b))
        System.out.print( "Yes");
    else
        System.out.print( "No");
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python 3 implementation of the
# above approach
N = 3
M = 3
 
# Boolean function that returns
# true or false
def check(a, b):
     
    # Traverse for all elements
    for i in range(1, N, 1):
        for j in range(1, M, 1):
             
            # If both are not equal
            if (a[i][j] != b[i][j]):
                 
                # Change the parity of
                # all corner elements
                a[i][j] ^= 1
                a[0][0] ^= 1
                a[0][j] ^= 1
                a[i][0] ^= 1
 
    # Check if A is equal to B
    for i in range(N):
        for j in range(M):
             
            # Not equal
            if (a[i][j] != b[i][j]):
                return False
     
    return True
 
# Driver Code
if __name__ == '__main__':
     
    # First binary matrix
    a = [[0, 1, 0],
         [0, 1, 0],
         [1, 0, 0]]
 
    # Second binary matrix
    b = [[1, 0, 0],
         [1, 0, 0],
         [1, 0, 0]]
 
    if (check(a, b)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of the
// above approach
using System;
 
class GFG
{
     
static readonly int N = 3,M =3;
 
// Boolean function that returns
// true or false
static bool check(int [,]a, int [,]b)
{
    // Traverse for all elements
    for (int i = 1; i < N; i++)
    {
        for (int j = 1; j < M; j++)
        {
 
            // If both are not equal
            if (a[i,j] != b[i,j])
            {
 
                // Change the parity of
                // all corner elements
                a[i,j] ^= 1;
                a[0,0] ^= 1;
                a[0,j] ^= 1;
                a[i,0] ^= 1;
            }
        }
    }
 
    // Check if A is equal to B
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
 
            // Not equal
            if (a[i,j] != b[i,j])
                return false;
        }
    }
    return true;
}
 
// Driver Code
public static void Main(String []args)
{
    // First binary matrix
    int [,]a = { { 0, 1, 0 },
                    { 0, 1, 0 },
                    { 1, 0, 0 } };
 
    // Second binary matrix
    int [,]b = { { 1, 0, 0 },
                    { 1, 0, 0 },
                    { 1, 0, 0 } };
 
    if (check(a, b))
        Console.Write( "Yes");
    else
        Console.Write( "No");
}
}
 
// This code has been contributed by 29AjayKumar

PHP




<?php
// PHP implementation of the above approach
 
$N = 3;
$M = 3 ;
 
// Boolean function that returns
// true or false
function check($a, $b)
{
    // Traverse for all elements
    for ($i = 1; $i < $GLOBALS['N']; $i++)
    {
        for ($j = 1; $j < $GLOBALS['M']; $j++)
        {
 
            // If both are not equal
            if ($a[$i][$j] != $b[$i][$j])
            {
 
                // Change the parity of
                // all corner elements
                $a[$i][$j] ^= 1;
                $a[0][0] ^= 1;
                $a[0][$j] ^= 1;
                $a[$i][0] ^= 1;
            }
        }
    }
 
    // Check if A is equal to B
    for ($i = 0; $i < $GLOBALS['N']; $i++)
    {
        for ($j = 0; $j < $GLOBALS['M']; $j++)
        {
 
            // Not equal
            if ($a[$i][$j] != $b[$i][$j])
                return false;
        }
    }
    return true;
}
 
    // Driver Code
    // First binary matrix
    $a = array(array( 0, 1, 0 ),
                array( 0, 1, 0 ),
                array( 1, 0, 0 ) );
 
    // Second binary matrix
    $b = array( array( 1, 0, 0 ),
                array(1, 0, 0 ),
                array( 1, 0, 0 ) );
 
    if (check($a, $b))
        echo "Yes";
    else
        echo "No";
 
// This code is contributed by Ryuga
?>

Javascript




<script>
// javascript implementation of the
// above approach    
var N = 3, M = 3;
 
    // Boolean function that returns
    // true or false
    function check(a , b) {
        // Traverse for all elements
        for (i = 1; i < N; i++) {
            for (j = 1; j < M; j++) {
 
                // If both are not equal
                if (a[i][j] != b[i][j]) {
 
                    // Change the parity of
                    // all corner elements
                    a[i][j] ^= 1;
                    a[0][0] ^= 1;
                    a[0][j] ^= 1;
                    a[i][0] ^= 1;
                }
            }
        }
 
        // Check if A is equal to B
        for (i = 0; i < N; i++) {
            for (j = 0; j < M; j++) {
 
                // Not equal
                if (a[i][j] != b[i][j])
                    return false;
            }
        }
        return true;
    }
 
    // Driver Code
     
        // First binary matrix
        var a = [ [ 0, 1, 0 ],
                [ 0, 1, 0 ],
                [ 1, 0, 0 ] ];
 
        // Second binary matrix
        var b = [ [ 1, 0, 0 ],
                [ 1, 0, 0 ],
                [ 1, 0, 0 ] ];
 
        if (check(a, b))
            document.write("Yes");
        else
            document.write("No");
 
// This code contributed by aashish1995
</script>

Output: 

Yes

 

Time Complexity: O(N * M)

Auxiliary Space: O(1)


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