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:
- 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++
#include <bits/stdc++.h>
using namespace std;
string check( int Arr[], int n,
int M, int K)
{
int flag = 0;
for ( int i = 0; i < n; i++) {
if (Arr[i] == 0)
flag = 1;
}
if (K % 2 != 0
&& flag == 0)
return "No" ;
else
return "Yes" ;
}
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;
}
|
Java
import java.util.*;
class GFG{
static String check( int []Arr, int n,
int M, int K)
{
int flag = 0 ;
for ( int i = 0 ; i < n; i++) {
if (Arr[i] == 0 )
flag = 1 ;
}
if (K % 2 != 0
&& flag == 0 )
return "No" ;
else
return "Yes" ;
}
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));
}
}
|
Python3
def check(Arr, n, M, K):
flag = 0
for i in range (n):
if (Arr[i] = = 0 ):
flag = 1
if (K % 2 ! = 0 and flag = = 0 ):
return "No"
else :
return "Yes" ;
if __name__ = = '__main__' :
Arr = [ 1 , 1 , 2 , 4 , 7 , 8 ]
M = 5 ;
K = 6 ;
n = len (Arr);
print (check(Arr, n, M, K))
|
C#
using System;
class GFG
{
static String check( int []Arr, int n, int M, int K)
{
int flag = 0;
for ( int i = 0; i < n; i++) {
if (Arr[i] == 0)
flag = 1;
}
if (K % 2 != 0
&& flag == 0)
return "No" ;
else
return "Yes" ;
}
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));
}
}
|
Javascript
<script>
function check(Arr, n, M, K)
{
let flag = 0;
for (let i = 0; i < n; i++) {
if (Arr[i] == 0)
flag = 1;
}
if (K % 2 != 0
&& flag == 0)
return "No" ;
else
return "Yes" ;
}
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>
|
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.
Last Updated :
29 May, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...