XOR of XORs of all sub-matrices
Given a ‘N*N’ matrix, the task is to find the XOR of XORs of all possible sub-matrices.
Examples:
Input :arr = {{3, 1}, {1, 3}} Output : 0 Explanation: All the elements lie in 4 submatrices each. 4 being even, there total contribution towards final answer becomes 0. Thus, ans = 0. Input : arr = {{6, 7, 13}, {8, 3, 4}, {9, 7, 6}}; Output : 4
A Simple Approach is to generate all the possible submatrices, find the XOR of each submatrix uniquely, and then XOR them all up. The time complexity of this approach will be O(n6).
Better solution: For each index(R, C), we will try to find the number of sub-matrices in which that index lies. If the number of sub-matrices is odd, then the final answer will be updated as ans = (ans ^ arr[R][C]). In case of even, we don’t need to update the answer. This works because a number XORed with itself gives zero and the order of operation doesn’t affect the final XOR value.
Assuming 0-based indexing, the number of sub-matrices an index (R, C) lies in equals
(R + 1)*(C + 1)*(N - R)*(N - C)
Below is the implementation of the above approach:
C++
// C++ program to find the XOR of XOR's of // all submatrices #include <iostream> using namespace std; #define n 3 // Function to find to required // XOR value int submatrixXor( int arr[][n]) { int ans = 0; // Nested loop to find the // number of sub-matrix each // index belongs to for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { // Number of ways to choose // from top-left elements int top_left = (i + 1) * (j + 1); // Number of ways to choose // from bottom-right elements int bottom_right = (n - i) * (n - j); if ((top_left % 2 == 1) && (bottom_right % 2 == 1)) ans = (ans ^ arr[i][j]); } } return ans; } // Driver Code int main() { int arr[][n] = { { 6, 7, 13 }, { 8, 3, 4 }, { 9, 7, 6 } }; cout << submatrixXor(arr); return 0; } |
Java
//Java program to find the XOR of XOR's // of all submatrices class GFG { // Function to find to required // XOR value static int submatrixXor( int [][]arr) { int n = 3 ; int ans = 0 ; // Nested loop to find the // number of sub-matrix each // index belongs to for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < n; j++) { // Number of ways to choose // from top-left elements int top_left = (i + 1 ) * (j + 1 ); // Number of ways to choose // from bottom-right elements int bottom_right = (n - i) * (n - j); if ((top_left % 2 == 1 ) && (bottom_right % 2 == 1 )) ans = (ans ^ arr[i][j]); } } return ans; } // Driver Code public static void main(String[] args) { int [][] arr = {{ 6 , 7 , 13 }, { 8 , 3 , 4 }, { 9 , 7 , 6 }}; System.out.println(submatrixXor(arr)); } } // This code is contributed // by Code_Mech. |
Python3
# Python3 program to find the XOR of # XOR's of all submatrices # Function to find to required # XOR value def submatrixXor(arr, n): ans = 0 # Nested loop to find the # number of sub-matrix each # index belongs to for i in range ( 0 , n): for j in range ( 0 , n): # Number of ways to choose # from top-left elements top_left = (i + 1 ) * (j + 1 ) # Number of ways to choose # from bottom-right elements bottom_right = (n - i) * (n - j) if (top_left % 2 = = 1 and bottom_right % 2 = = 1 ): ans = (ans ^ arr[i][j]) return ans # Driver code n = 3 arr = [[ 6 , 7 , 13 ], [ 8 , 3 , 4 ], [ 9 , 7 , 6 ]] print (submatrixXor(arr, n)) # This code is contributed by Shrikant13 |
C#
// C# program to find the XOR of XOR's // of all submatrices using System; class GFG { // Function to find to required // XOR value static int submatrixXor( int [,]arr) { int n = 3; int ans = 0; // Nested loop to find the // number of sub-matrix each // index belongs to for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { // Number of ways to choose // from top-left elements int top_left = (i + 1) * (j + 1); // Number of ways to choose // from bottom-right elements int bottom_right = (n - i) * (n - j); if ((top_left % 2 == 1) && (bottom_right % 2 == 1)) ans = (ans ^ arr[i, j]); } } return ans; } // Driver Code public static void Main() { int [, ]arr = {{ 6, 7, 13}, { 8, 3, 4 }, { 9, 7, 6 }}; Console.Write(submatrixXor(arr)); } } // This code is contributed // by Akanksha Rai |
PHP
<?php // PHP program to find the XOR of // XOR's of all submatrices // Function to find to required // XOR value function submatrixXor( $arr ) { $ans = 0; $n = 3 ; // Nested loop to find the // number of sub-matrix each // index belongs to for ( $i = 0; $i < $n ; $i ++) { for ( $j = 0; $j < $n ; $j ++) { // Number of ways to choose // from top-left elements $top_left = ( $i + 1) * ( $j + 1); // Number of ways to choose // from bottom-right elements $bottom_right = ( $n - $i ) * ( $n - $j ); if (( $top_left % 2 == 1) && ( $bottom_right % 2 == 1)) $ans = ( $ans ^ $arr [ $i ][ $j ]); } } return $ans ; } // Driver Code $arr = array ( array ( 6, 7, 13 ), array ( 8, 3, 4 ), array ( 9, 7, 6 )); echo submatrixXor( $arr ); # This code is contributed by Ryuga ?> |
Javascript
<script> // Javascript program to find the // XOR of XOR's of // all submatrices const n = 3; // Function to find to required // XOR value function submatrixXor(arr) { let ans = 0; // Nested loop to find the // number of sub-matrix each // index belongs to for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) { // Number of ways to choose // from top-left elements let top_left = (i + 1) * (j + 1); // Number of ways to choose // from bottom-right elements let bottom_right = (n - i) * (n - j); if ((top_left % 2 == 1) && (bottom_right % 2 == 1)) ans = (ans ^ arr[i][j]); } } return ans; } // Driver Code let arr = [ [ 6, 7, 13 ], [ 8, 3, 4 ], [ 9, 7, 6 ] ]; document.write(submatrixXor(arr)); </script> |
4
Time Complexity: O(N2)
Auxiliary Space: O(1)
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