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Maximum XOR value in matrix
• Difficulty Level : Easy
• Last Updated : 30 Apr, 2021

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 .

1. 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.
2. Store the maximum of both.

Below is the implementation :

## C++

 `// C++ program to Find maximum XOR value in``// matrix either row / column wise``#include``using` `namespace` `std;` `// maximum number of row and column``const` `int` `MAX = 1000;` `// function return the maximum xor value that is``// either row or column wise``int` `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 Code``int` `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 wise``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 column``MAX` `=` `1000`` ` `# Function return the maximum``# xor value that is either row``# or column wise``def` `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 Code``N ``=` `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 wise``using` `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

 ``

## Javascript

 ``

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 GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.