# Maximum XOR value in matrix

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
```

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

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

 ` `

Output :

```maximum XOR value  : 12
```

Time complexity : O(N*N)
space complexity : O(1)

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Improved By : nitin mittal, vt_m

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