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

Given two binary matrices A[][] and B[][] of N X 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 same or not. If they are the same, we can change A to B.

Below is the implementation of the above approach:

 `// C++ implementation of the ` `// above appoach ` ` `  `#include ` `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 implementation of the  ` `// above appoach  ` `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 `

 `# Python 3 implementation of the ` `# above appoach ` `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# implementation of the  ` `// above appoach  ` `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 `

 ` `

Output:
```Yes
```

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