Check if a number from every row can be selected such that xor of the numbers is greater than zero

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 <bits/stdc++.h>
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][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
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

<?php
// PHP 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
function check($mat)
{

    global $N ;
    global $M ;

    $xorr = 0;

    // Find the xor of first
    // column for every row
    for ($i = 0; $i < $N; $i++)
    {
        $xorr =     $xorr ^ $mat[$i][0];
    }

    // If Xorr is 0
    if ($xorr != 0)
        return true;

    // Traverse in the matrix
    for ($i = 0; $i < $N; $i++)
    {
        for ( $j = 1; $j < $M; $j++)
        {

            // Check is atleast
            // 2 distinct elements
            if ($mat[$i][$j] != $mat[$i][0])
                return true;
        }
    }
    return false;
}

    // Driver code
    $mat = array(array( 7, 7, 7 ),
                    array( 10, 10, 7 ));

    if (check($mat))
        echo "Yes";
    else
        echo "No";

// This code is contributed by Tushil..
?>

Javascript

<script>
// Javascript program to implement
// the above approach

let N = 2;
let M = 3;

// Function to check if a number from every row
// can be selected such that xor of the numbers
// is greater than zero
function check(mat)
{
    let xorr = 0;

    // Find the xor of first
    // column for every row
    for (let i = 0; i < N; i++) {
        xorr ^= mat[i][0];
    }

    // If Xorr is 0
    if (xorr != 0)
        return true;

    // Traverse in the matrix
    for (let i = 0; i < N; i++) {
        for (let j = 1; j < M; j++) {

            // Check is atleast
            // 2 distinct elements
            if (mat[i][j] != mat[i][0])
                return true;
        }
    }

    return false;
}

// Driver code
    let mat = [ [ 7, 7, 7 ],
                      [ 10, 10, 7 ] ];

    if (check(mat))
        document.write("Yes");
    else
        document.write("No");

// This code is contributed by souravmahato348.
</script>

Output:

Yes

Time Complexity: O(N * M)

Auxiliary Space: O(1)