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:

- if even number of XOR operations are performed for any positive number, then the original number can be retained.
- 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.
- Therefore,
**if K is even and M is 0**, then the answer will always be Yes. **If K is odd and 0 is not present in the array**, then the answer will always be No.**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++

`// 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; ` `} ` |

*chevron_right*

*filter_none*

## Java

`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 ` |

*chevron_right*

*filter_none*

## Python3

`# 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 ` |

*chevron_right*

*filter_none*

## C#

`// 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 ` |

*chevron_right*

*filter_none*

**Output:**

Yes

## Recommended Posts:

- Check if the last element of array is even or odd after performing a operation p times
- Check if at least half array is reducible to zero by performing some operations
- Maximum inversions in a sequence of 1 to N after performing given operations at most K times
- Check whether K times of a element is present in array
- Check if given array can be made 0 with given operations performed any number of times
- Check if the given array can be reduced to zeros with the given operation performed given number of times
- Find original array from encrypted array (An array of sums of other elements)
- Generate original array from an array that store the counts of greater elements on right
- Count number of permutation of an Array having no SubArray of size two or more from original Array
- Rearrange an Array such that Sum of same-indexed subsets differ from their Sum in the original Array
- Print n smallest elements from given array in their original order
- Find the deleted value from the array when average of original elements is given
- Find the original Array using XOR values of all adjacent elements
- Generate original array from difference between every two consecutive elements
- Maximum possible Array sum after performing given operations
- Find the sum of array after performing every query
- Sum of the updated array after performing the given operation
- Maximum possible array sum after performing the given operation
- Count subarrays having total distinct elements same as original array
- Minimum time to return array to its original state after given modifications

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

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.