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 timesstring 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 Codeint 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;} |
Java
import java.util.*;class GFG{// Function to check if original Array// can be retained by performing XOR// with M exactly K timesstatic 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 Codepublic 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
# Python3 implementation for the# above mentioned problem # Function to check if original Array# can be retained by performing XOR# with M exactly K timesdef 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 Codeif __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#
// C# implementation for the// above mentioned problemusing System; class GFG { // Function to check if original Array// can be retained by performing XOR// with M exactly K timesstatic 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 |
Javascript
<script>// Javascript implementation for the// above mentioned problem// Function to check if original Array// can be retained by performing XOR// with M exactly K timesfunction 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> |
Output:
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.
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