Check if original Array is retained after performing XOR with M exactly K times

Given an array A and two integers M and K, the task is to check and print "Yes", if the original array can be retained by performing exactly 'K' number of bitwise XOR operations of the array elements with 'M'. Else print "No".
Note: XOR operation can be performed, on any element of the array, for 0 or more times. 
Examples: 
 

Input: A[] = {1, 2, 3, 4}, M = 5, K = 6 
Output: Yes 
Explanation: 
If the XOR is performed on 1st element, A[0], for 6 times, we get A[0] back. Therefore, the original array is retained.
Input: A[] = {5, 9, 3, 4, 5}, M = 5, K = 3 
Output: No 
Explanation: 
The original array cant be retained after performing odd number of XOR operations. 
 

Approach: This problem can be solved using the XOR property 
 

A XOR B = C and C XOR B = A 
 

It can be seen that: 
 

1. if even number of XOR operations are performed for any positive number, then the original number can be retained.
2. However, 0 is an exception. If both odd or even number of XOR operations are performed for 0, then the original number can be retained.
3. Therefore, if K is even and M is 0, then the answer will always be Yes.
4. If K is odd and 0 is not present in the array, then the answer will always be No.
5. If K is odd and the count of 0 is at least 1 in the array then, the answer will be Yes.

Below is the implementation of the above approach:
 

// C++ implementation for the
// above mentioned problem

#include <bits/stdc++.h>
using namespace std;

// Function to check if original Array
// can be retained by performing XOR
// with M exactly K times
string check(int Arr[], int n,
             int M, int K)
{
    int flag = 0;

    // Check if O is present or not
    for (int i = 0; i < n; i++) {
        if (Arr[i] == 0)
            flag = 1;
    }

    // If K is odd and 0 is not present
    // then the answer will always be No.
    if (K % 2 != 0
        && flag == 0)
        return "No";

    // Else it will be Yes
    else
        return "Yes";
}

// Driver Code
int main()
{

    int Arr[] = { 1, 1, 2, 4, 7, 8 };
    int M = 5;
    int K = 6;
    int n = sizeof(Arr) / sizeof(Arr[0]);

    cout << check(Arr, n, M, K);
    return 0;
}

import java.util.*;

class GFG{

// Function to check if original Array
// can be retained by performing XOR
// with M exactly K times
static String check(int []Arr, int n,
            int M, int K)
{
    int flag = 0;

    // Check if O is present or not
    for (int i = 0; i < n; i++) {
        if (Arr[i] == 0)
            flag = 1;
    }

    // If K is odd and 0 is not present
    // then the answer will always be No.
    if (K % 2 != 0
        && flag == 0)
        return "No";

    // Else it will be Yes
    else
        return "Yes";
}

// Driver Code
public static void main(String args[])
{

    int []Arr = { 1, 1, 2, 4, 7, 8 };
    int M = 5;
    int K = 6;
    int n = Arr.length;

    System.out.println(check(Arr, n, M, K));
}
}

// This code is contributed by Surendra_Gangwar

# Python3 implementation for the
# above mentioned problem
 
# Function to check if original Array
# can be retained by performing XOR
# with M exactly K times
def check(Arr,  n, M,  K):
    flag = 0
 
    # Check if O is present or not
    for i in range(n):
        if (Arr[i] == 0):
            flag = 1
    
    # If K is odd and 0 is not present
    # then the answer will always be No.
    if (K % 2 != 0 and flag == 0):
        return "No"
 
    # Else it will be Yes
    else:
        return "Yes";
 
# Driver Code
if __name__=='__main__': 

    Arr = [ 1, 1, 2, 4, 7, 8 ]
    M = 5;
    K = 6;
    n = len(Arr);
 
    print(check(Arr, n, M, K))

# This article contributed by Princi Singh

// C# implementation for the
// above mentioned problem
using System; 
  
class GFG 
{ 
  

// Function to check if original Array
// can be retained by performing XOR
// with M exactly K times
static String check(int []Arr, int n,int M, int K)
{
    int flag = 0;

    // Check if O is present or not
    for (int i = 0; i < n; i++) {
        if (Arr[i] == 0)
            flag = 1;
    }

    // If K is odd and 0 is not present
    // then the answer will always be No.
    if (K % 2 != 0
        && flag == 0)
        return "No";

    // Else it will be Yes
    else
        return "Yes";
}

// Driver code 
public static void Main(String[] args) 
{

    int []Arr = { 1, 1, 2, 4, 7, 8 };
    int M = 5;
    int K = 6;
    int n = Arr.Length; 

    Console.Write(check(Arr, n, M, K));
}
}

// This code is contributed by shivanisinghss2110

<script>
// Javascript implementation for the
// above mentioned problem

// Function to check if original Array
// can be retained by performing XOR
// with M exactly K times
function check(Arr, n, M, K)
{
    let flag = 0;

    // Check if O is present or not
    for (let i = 0; i < n; i++) {
        if (Arr[i] == 0)
            flag = 1;
    }

    // If K is odd and 0 is not present
    // then the answer will always be No.
    if (K % 2 != 0
        && flag == 0)
        return "No";

    // Else it will be Yes
    else
        return "Yes";
}

// Driver Code

    let Arr = [ 1, 1, 2, 4, 7, 8 ];
    let M = 5;
    let K = 6;
    let n = Arr.length;

    document.write(check(Arr, n, M, K));

</script>

Yes

Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time 
Auxiliary Space: O(1), as we are not using any extra space.