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

• Last Updated : 06 Sep, 2021

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

Examples:

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Input: N = 123, K = 3
Output: 123
(123 ^ 123 ^ 123) = 123
Input: N = 123, K = 2
Output:
(123 ^ 123) = 0

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

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``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` `res;``}` `// Driver code``int` `main()``{``    ``int` `n = 123, k = 3;` `    ``cout << xorK(n, k);` `    ``return` `0;``}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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.`

## Javascript

 ``
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++

 `// C++ implementation of the approach``#include ``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;``}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

## Javascript

 ``
Output:
`123`

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

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