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Check if a number from every row can be selected such that xor of the numbers is greater than zero
• Last Updated : 27 Apr, 2021

Given a 2-D array of order N X M array elements, the task is to check if we can select a number from every row in such a way that xor of the selected numbers is greater than 0
Note: There is a minimum of 2 rows.
Examples:

```Input: a[][] = {{7, 7, 7},
{10, 10, 7}}
Output: Yes

Input: a[][] = {{1, 1, 1},
{1, 1, 1},
{1, 1, 1},
{1, 1, 1}}
Output: No ```

Approach: Initially check if xor of first column elements of every row is 0 or not. If it is non-zero then it is possible. If it is zero, check if any of the rows has two or more distinct elements, then also it is possible. If both of the above conditions are not satisfied, then it is not possible.
Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;``#define N 2``#define M 3` `// Function to check if a number from every row``// can be selected such that xor of the numbers``// is greater than zero``bool` `check(``int` `mat[N][M])``{``    ``int` `xorr = 0;` `    ``// Find the xor of first``    ``// column for every row``    ``for` `(``int` `i = 0; i < N; i++) {``        ``xorr ^= mat[i];``    ``}` `    ``// If Xorr is 0``    ``if` `(xorr != 0)``        ``return` `true``;` `    ``// Traverse in the matrix``    ``for` `(``int` `i = 0; i < N; i++) {``        ``for` `(``int` `j = 1; j < M; j++) {` `            ``// Check is atleast``            ``// 2 distinct elements``            ``if` `(mat[i][j] != mat[i])``                ``return` `true``;``        ``}``    ``}` `    ``return` `false``;``}` `// Driver code``int` `main()``{``    ``int` `mat[N][M] = { { 7, 7, 7 },``                      ``{ 10, 10, 7 } };` `    ``if` `(check(mat))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.io.*;` `class` `GFG``{``    ``static` `int` `N = ``2``;``    ``static` `int` `M = ``3``;``    ` `    ``// Function to check if a number``    ``// from every row can be selected``    ``// such that xor of the numbers``    ``// is greater than zero``    ``static` `boolean` `check(``int` `mat[][])``    ``{``        ``int` `xorr = ``0``;``    ` `        ``// Find the xor of first``        ``// column for every row``        ``for` `(``int` `i = ``0``; i < N; i++)``        ``{``            ``xorr ^= mat[i] [``0``];``        ``}``    ` `        ``// If Xorr is 0``        ``if` `(xorr != ``0``)``            ``return` `true``;``    ` `        ``// Traverse in the matrix``        ``for` `(``int` `i = ``0``; i < N; i++)``        ``{``            ``for` `(``int` `j = ``1``; j < M; j++)``            ``{``    ` `                ``// Check is atleast``                ``// 2 distinct elements``                ``if` `(mat[i] [j] != mat[i] [``0``])``                    ``return` `true``;``            ``}``        ``}``    ` `        ``return` `false``;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ` `        ``int` `mat[][] = {{ ``7``, ``7``, ``7` `},``                    ``{ ``10``, ``10``, ``7` `}};``    ` `        ``if` `(check(mat))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);` `    ``}``}` `// This code is contributed by ajit`

## Python3

 `# Python3 program to implement``# the above approach``N ``=` `2``M ``=` `3` `# Function to check if a number from every row``# can be selected such that xor of the numbers``# is greater than zero``def` `check(mat):` `    ``xorr ``=` `0` `    ``# Find the xor of first``    ``# column for every row``    ``for` `i ``in` `range``(N):``        ``xorr ^``=` `mat[i][``0``]` `    ``# If Xorr is 0``    ``if` `(xorr !``=` `0``):``        ``return` `True` `    ``# Traverse in the matrix``    ``for` `i ``in` `range``(N):``        ``for` `j ``in` `range``(``1``, M):` `            ``# Check is atleast``            ``# 2 distinct elements``            ``if` `(mat[i][j] !``=` `mat[i][``0``]):``                ``return` `True``        ` `    ``return` `False` `# Driver code``mat ``=` `[[ ``7``, ``7``, ``7` `],``       ``[ ``10``, ``10``, ``7` `]]` `if` `(check(mat)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# This code is contributed by mohit kumar`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG``{``    ``static` `int` `N = 2;``    ``static` `int` `M = 3;``    ` `    ``// Function to check if a number``    ``// from every row can be selected``    ``// such that xor of the numbers``    ``// is greater than zero``    ``static` `bool` `check(``int` `[,]mat)``    ``{``        ``int` `xorr = 0;``    ` `        ``// Find the xor of first``        ``// column for every row``        ``for` `(``int` `i = 0; i < N; i++)``        ``{``            ``xorr ^= mat[i, 0];``        ``}``    ` `        ``// If Xorr is 0``        ``if` `(xorr != 0)``            ``return` `true``;``    ` `        ``// Traverse in the matrix``        ``for` `(``int` `i = 0; i < N; i++)``        ``{``            ``for` `(``int` `j = 1; j < M; j++)``            ``{``    ` `                ``// Check is atleast``                ``// 2 distinct elements``                ``if` `(mat[i, j] != mat[i, 0])``                    ``return` `true``;``            ``}``        ``}``    ` `        ``return` `false``;``    ``}``    ` `    ``// Driver code``    ``static` `void` `Main()``    ``{``        ``int` `[,]mat = {{ 7, 7, 7 },``                      ``{ 10, 10, 7 }};``    ` `        ``if` `(check(mat))``            ``Console.Write(``"Yes"``);``        ``else``            ``Console.Write(``"No"``);``    ``}``}` `// This code is contributed by mits`

## PHP

 ``

## Javascript

 ``
Output:
`Yes`

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