Find the value of N XOR’ed to itself K times

Given two integer N and K, the task is to find the value of N XOR N XOR N XOR … XOR N (K times).

Examples:

Input: N = 123, K = 3
Output: 123
(123 ^ 123 ^ 123) = 123

Input: N = 123, K = 2
Output: 0
(123 ^ 123) = 0

Naive approach: Simply run a for loop and xor N, K times.



Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return n ^ n ^ ... k times
int xorK(int n, int k)
{
  
    // Find the result
    int res = n;
    for (int i = 1; i < k; i++)
        res = (res ^ n);
    return n;
}
  
// Driver code
int main()
{
    int n = 123, k = 3;
  
    cout << xorK(n, k);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
  
// Function to return n ^ n ^ ... k times
static int xorK(int n, int k)
{
  
    // Find the result
    int res = n;
    for (int i = 1; i < k; i++)
        res = (res ^ n);
    return n;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 123, k = 3;
  
    System.out.print(xorK(n, k));
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 implementation of the approach
  
# Function to return n ^ n ^ ... k times
def xorK(n, k):
  
    # Find the result
    res = n
    for i in range(1, k):
        res = (res ^ n)
    return n
  
# Driver code
n = 123
k = 3
  
print(xorK(n, k))
  
# This code is contributed by Mohit Kumar

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return n ^ n ^ ... k times
static int xorK(int n, int k)
{
  
    // Find the result
    int res = n;
    for (int i = 1; i < k; i++)
        res = (res ^ n);
    return n;
}
  
// Driver code
static public void Main ()
{
    int n = 123, k = 3;
      
    Console.Write(xorK(n, k));
}
}
  
// This code is contributed by ajit.

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Output:

123

Time Complexity: O(K)
Space Complexity: O(1)

Efficient approach: From the properties X XOR X = 0 and X ^ 0 = X, it can be observed that when K is odd then the answer will be N itself else the answer will be 0.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return n ^ n ^ ... k times
int xorK(int n, int k)
{
  
    // If k is odd the answer is
    // the number itself
    if (k % 2 == 1)
        return n;
  
    // Else the answer is 0
    return 0;
}
  
// Driver code
int main()
{
    int n = 123, k = 3;
  
    cout << xorK(n, k);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
  
// Function to return n ^ n ^ ... k times
static int xorK(int n, int k)
{
  
    // If k is odd the answer is
    // the number itself
    if (k % 2 == 1)
        return n;
  
    // Else the answer is 0
    return 0;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 123, k = 3;
  
    System.out.print(xorK(n, k));
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python implementation of the approach 
  
# Function to return n ^ n ^ ... k times 
def xorK(n, k):
      
    # If k is odd the answer is 
    # the number itself 
    if (k % 2 == 1):
        return n
  
    # Else the answer is 0 
    return 0
  
# Driver code 
n = 123
k = 3
  
print(xorK(n, k))
  
# This code is contributed by Sanjit_Prasad

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return n ^ n ^ ... k times
static int xorK(int n, int k)
{
  
    // If k is odd the answer is
    // the number itself
    if (k % 2 == 1)
        return n;
  
    // Else the answer is 0
    return 0;
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 123, k = 3;
  
    Console.Write(xorK(n, k));
}
}
  
// This code is contributed by 29AjayKumar

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Output:

123

Time Complexity: O(1)
Space Complexity: O(1)

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