Given a square matrix (N X N), the task is to find the maximum XOR value of a complete row or a complete column.
Examples :
Input : N = 3
mat[3][3] = {{1, 0, 4},
{3, 7, 2},
{5, 9, 10} };
Output : 14
We get this maximum XOR value by doing XOR
of elements in second column 0 ^ 7 ^ 9 = 14
Input : N = 4
mat[4][4] = { {1, 2, 3, 6},
{4, 5, 6,7},
{7, 8, 9, 10},
{2, 4, 5, 11}}
Output : 12
A simple solution of this problem is we can traverse the matrix twice and calculate maximum xor value row-wise & column wise ,and at last return the maximum between (xor_row , xor_column).
A efficient solution is we can traverse matrix only one time and calculate max XOR value .
- Start traverse the matrix and calculate XOR at each index row and column wise. We can compute both values by using indexes in reverse way. This is possible because matrix is a square matrix.
- Store the maximum of both.
Below is the implementation :
C++
// C++ program to Find maximum XOR value in// matrix either row / column wise#include<iostream>using namespace std;// maximum number of row and columnconst int MAX = 1000;// function return the maximum xor value that is// either row or column wiseint maxXOR(int mat[][MAX], int N){ // for row xor and column xor int r_xor, c_xor; int max_xor = 0; // traverse matrix for (int i = 0 ; i < N ; i++) { r_xor = 0, c_xor = 0; for (int j = 0 ; j < N ; j++) { // xor row element r_xor = r_xor^mat[i][j]; // for each column : j is act as row & i // act as column xor column element c_xor = c_xor^mat[j][i]; } // update maximum between r_xor , c_xor if (max_xor < max(r_xor, c_xor)) max_xor = max(r_xor, c_xor); } // return maximum xor value return max_xor;}// driver Codeint main(){ int N = 3; int mat[][MAX] = {{1 , 5, 4}, {3 , 7, 2 }, {5 , 9, 10} }; cout << "maximum XOR value : " << maxXOR(mat, N); return 0;} |
Java
// Java program to Find maximum XOR value in// matrix either row / column wiseimport java.io.*;class GFG { // maximum number of row and column static final int MAX = 1000; // function return the maximum xor value // that is either row or column wise static int maxXOR(int mat[][], int N) { // for row xor and column xor int r_xor, c_xor; int max_xor = 0; // traverse matrix for (int i = 0 ; i < N ; i++) { r_xor = 0; c_xor = 0; for (int j = 0 ; j < N ; j++) { // xor row element r_xor = r_xor^mat[i][j]; // for each column : j is act as row & i // act as column xor column element c_xor = c_xor^mat[j][i]; } // update maximum between r_xor , c_xor if (max_xor < Math.max(r_xor, c_xor)) max_xor = Math.max(r_xor, c_xor); } // return maximum xor value return max_xor; } //driver code public static void main (String[] args) { int N = 3; int mat[][] = { {1, 5, 4}, {3, 7, 2}, {5, 9, 10}}; System.out.print("maximum XOR value : " + maxXOR(mat, N)); }}// This code is contributed by Anant Agarwal. |
Python3
# Python3 program to Find maximum# XOR value in matrix either row / column wise# maximum number of row and columnMAX = 1000 # Function return the maximum# xor value that is either row# or column wisedef maxXOR(mat, N): # For row xor and column xor max_xor = 0 # Traverse matrix for i in range(N): r_xor = 0 c_xor = 0 for j in range(N): # xor row element r_xor = r_xor ^ mat[i][j] # for each column : j is act as row & i # act as column xor column element c_xor = c_xor ^ mat[j][i] # update maximum between r_xor , c_xor if (max_xor < max(r_xor, c_xor)): max_xor = max(r_xor, c_xor) # return maximum xor value return max_xor # Driver CodeN = 3 mat= [[1 , 5, 4], [3 , 7, 2 ], [5 , 9, 10]] print("maximum XOR value : ", maxXOR(mat, N)) # This code is contributed by Anant Agarwal. |
C#
// C# program to Find maximum XOR value in// matrix either row / column wiseusing System;class GFG { // maximum number of row and column // function return the maximum xor value // that is either row or column wise static int maxXOR(int [,]mat, int N) { // for row xor and column xor int r_xor, c_xor; int max_xor = 0; // traverse matrix for (int i = 0 ; i < N ; i++) { r_xor = 0; c_xor = 0; for (int j = 0 ; j < N ; j++) { // xor row element r_xor = r_xor^mat[i, j]; // for each column : j is act as row & i // act as column xor column element c_xor = c_xor^mat[j, i]; } // update maximum between r_xor , c_xor if (max_xor < Math.Max(r_xor, c_xor)) max_xor = Math.Max(r_xor, c_xor); } // return maximum xor value return max_xor; } // Driver code public static void Main () { int N = 3; int [,]mat = { {1, 5, 4}, {3, 7, 2}, {5, 9, 10}}; Console.Write("maximum XOR value : " + maxXOR(mat, N)); }}// This code is contributed by nitin mittal. |
PHP
<?php// PHP program to Find // maximum XOR value in// matrix either row or // column wise// maximum number of // row and column$MAX = 1000;// function return the maximum // xor value that is either// row or column wisefunction maxXOR($mat, $N){ // for row xor and // column xor $r_xor; $c_xor; $max_xor = 0; // traverse matrix for ($i = 0 ; $i < $N ; $i++) { $r_xor = 0; $c_xor = 0; for ($j = 0 ; $j < $N ; $j++) { // xor row element $r_xor = $r_xor^$mat[$i][$j]; // for each column : j // is act as row & i // act as column xor // column element $c_xor = $c_xor^$mat[$j][$i]; } // update maximum between // r_xor , c_xor if ($max_xor < max($r_xor, $c_xor)) $max_xor = max($r_xor, $c_xor); } // return maximum // xor value return $max_xor;} // Driver Code $N = 3; $mat = array(array(1, 5, 4), array(3, 7, 2), array(5, 9, 10)); echo "maximum XOR value : " , maxXOR($mat, $N);// This code is contributed by anuj_67.?> |
Javascript
<script>// Javascript program to Find// maximum XOR value in// matrix either row / column wise// maximum number of row and columnconst MAX = 1000;// function return the maximum // xor value that is// either row or column wisefunction maxXOR(mat, N){ // for row xor and column xor let r_xor, c_xor; let max_xor = 0; // traverse matrix for (let i = 0 ; i < N ; i++) { r_xor = 0, c_xor = 0; for (let j = 0 ; j < N ; j++) { // xor row element r_xor = r_xor^mat[i][j]; // for each column : j // is act as row & i // act as column xor // column element c_xor = c_xor^mat[j][i]; } // update maximum between r_xor , c_xor if (max_xor < Math.max(r_xor, c_xor)) max_xor = Math.max(r_xor, c_xor); } // return maximum xor value return max_xor;}// driver Code let N = 3; let mat = [[1 , 5, 4], [3 , 7, 2 ], [5 , 9, 10]]; document.write("maximum XOR value : " + maxXOR(mat, N));</script> |
Output
maximum XOR value : 12
Time complexity : O(N*N)
space complexity : O(1)
This article is contributed by Nishant_Singh(Pintu). If you like neveropen and would like to contribute, you can also write an article using write.neveropen.co.za or mail your article to review-team@neveropen.co.za. See your article appearing on the neveropen main page and help other Geeks.
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