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Sum of array elements after every element x is XORed itself x times

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  • Difficulty Level : Easy
  • Last Updated : 21 Jul, 2022
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Given an array of integers, the task is to compute sum of all array elements after doing XOR of each element x with itself x times. For example, if element is 4 so we do XOR of this number with itself 4 time Like:= 4^4^4^4 

Examples: 

Input :  arr[] = { 1, 2, 3, 5 }
Output :  9 
explanation:  1 + 2^2 + 3^3^3 + 5^5^5^5^5 : 9
                     
Input :   arr[] ={ 5, 6, 7, 9 }
Output :  21

A Simple solution is to pick each array element one by one and do its XOR with itself according to the its value. Finally add XOR values to the result. 

Below is the implementation of above idea. 

C++




// C++ program to compute sum of all element after
// doing Xor with itself ( element_time)
#include <bits/stdc++.h>
using namespace std;
 
// function return sum of all XOR element of array
int XorSum(int arr[], int n)
{
    // store result
    int result = 0;
 
    // Traverse array element and apply XOR
    // operation on it
    for (int i = 0; i < n; i++) {
         
        // XOR of current element with itself
        // according to value.
        int k = 0;
        for (int j = 1; j <= arr[i]; j++)
            k ^= arr[i];
 
        result += k;
    }
    return result;
}
// Driver program
int main()
{
    int arr[] = { 1, 2, 6, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << XorSum(arr, n) << endl;
    return 0;
}

Java




// Java program to compute sum of all
// element after doing Xor with itself
// ( element_time)
import java.io.*;
 
class GFG {
     
    // function return sum of all XOR
    // element of array
    static int XorSum(int arr[], int n)
    {
        // store result
        int result = 0;
     
        // Traverse array element and apply
        // XOR operation on it
        for (int i = 0; i < n; i++) {
             
            // XOR of current element with
            // itself according to value.
            int k = 0;
            for (int j = 1; j <= arr[i]; j++)
                k ^= arr[i];
     
            result += k;
        }
         
        return result;
    }
     
    // Driver program
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 6, 3, 4, 5 };
        int n = arr.length;
        System.out.println(XorSum(arr, n));
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

Python




# Python 3 program to compute sum of
# all element after doing Xor with
# itself ( element_time)
 
# function return sum of all XOR
# element of array
def XorSum(arr, n) :
     
    # store result
    result = 0
 
    # Traverse array element and
    # apply XOR operation on it
    for i in range(0, n) :
         
        # XOR of current element
        # with itself according to
        # value.
        k = 0
        for j in range(1, arr[i]+1) :
            k = k ^ arr[i]
 
        result = result + k
     
    return result
 
 
# Driver program
 
arr = [ 1, 2, 6, 3, 4, 5 ]
n = len(arr)
print(XorSum(arr, n))
 
 
# This code is contributed by Nikita Tiwari.

C#




// C# program to compute sum of all
// element after doing Xor with itself
// ( element_time)
using System;
 
class GFG {
     
    // function return sum of all XOR
    // element of array
    static int XorSum(int []arr, int n)
    {
        // store result
        int result = 0;
     
        // Traverse array element and apply
        // XOR operation on it
        for (int i = 0; i < n; i++) {
             
            // XOR of current element with
            // itself according to value.
            int k = 0;
            for (int j = 1; j <= arr[i]; j++)
                k ^= arr[i];
     
            result += k;
        }
         
        return result;
    }
     
    // Driver program
    public static void Main()
    {
        int []arr = { 1, 2, 6, 3, 4, 5 };
        int n = arr.Length;
        Console.WriteLine(XorSum(arr, n));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to compute
// sum of all element after
// doing Xor with itself
// ( element_time)
 
// function return sum of all
// XOR element of array
function XorSum( $arr, $n)
{
     
    // store result
    $result = 0;
 
    // Traverse array element
    // and apply XOR
    // operation on it
    for ($i = 0; $i < $n; $i++)
    {
         
        // XOR of current element
        // with itself according
        // to value.
        $k = 0;
        for ($j = 1; $j <= $arr[$i]; $j++)
            $k ^= $arr[$i];
 
        $result += $k;
    }
    return $result;
}
 
    // Driver Code
    $arr = array(1, 2, 6, 3, 4, 5);
    $n = count($arr);
    echo XorSum($arr, $n);
     
// This code is contributed by anuj_67.
?>

Javascript




<script>
// Javascript program to compute sum of all element after
// doing Xor with itself ( element_time)
 
// function return sum of all XOR element of array
function XorSum(arr, n)
{
    // store result
    let result = 0;
 
    // Traverse array element and apply XOR
    // operation on it
    for (let i = 0; i < n; i++) {
         
        // XOR of current element with itself
        // according to value.
        let k = 0;
        for (let j = 1; j <= arr[i]; j++)
            k ^= arr[i];
 
        result += k;
    }
    return result;
}
// Driver program
    let arr = [ 1, 2, 6, 3, 4, 5 ];
    let n = arr.length;
    document.write(XorSum(arr, n));
 
</script>

Output

9

Time Complexity: O(n*m) (here m is the maximum element in array ) 
Auxiliary Space: O(1)

Efficient solution of this problem is based on the fact that if we do a XOR of any number with itself( even number of times) it produces 0 and if we do a XOR odd number of time it produces same number. 

For Example 

   let number be  : 3  do XOR with itself 3 time
             3^3^3 = 3 
   let number be :  4 do XOR with itself 4 time 
             4^4^4^4 = 0 
  so if number is odd it's mean output is number 
  itself. Else zero  

Below is the implementation of above idea : 

C++




// C++ program to compute sum of all element after
// doing XOR with itself ( element_time)
#include <bits/stdc++.h>
using namespace std;
 
// function return sum of all XOR element of array
int XorSum(int arr[], int n)
{
    int result = 0;
    for (int i = 0; i < n; i++) {
 
        // if number is odd then add it to the
        // result else not
        if (arr[i] % 2 != 0)
            result += arr[i];
    }
 
    return result;
}
 
// Driver program
int main()
{
    int arr[] = { 1, 2, 6, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << XorSum(arr, n) << endl;
    return 0;
}

Java




// Java program to compute sum of
// all element after doing XOR
// with itself ( element_time)
class GFG {
     
// function return sum of all
// XOR element of array
static int XorSum(int arr[], int n) {
     
    int result = 0;
    for (int i = 0; i < n; i++) {
 
    // if number is odd then add it to the
    // result else not
    if (arr[i] % 2 != 0)
        result += arr[i];
    }
 
    return result;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = {1, 2, 6, 3, 4, 5};
    int n = arr.length;
    System.out.println(XorSum(arr, n));
}
}
 
// This code is contributed by Anant Agarwal.

Python3




# Python program to compute
# sum of all element after
# doing XOR with itself
# ( element_time)
 
# function return sum of
# all XOR element of array
def XorSum(arr,n):
 
    result = 0
    for i in range(n):
  
        # if number is odd then add it to the
        # result else not
        if (arr[i] % 2 != 0):
            result += arr[i]
     
  
    return result
 
# Driver program
arr = [ 1, 2, 6, 3, 4, 5 ]
n = len(arr)
 
print(XorSum(arr, n))
 
# This code is contributed
# by Anant Agarwal.

C#




// C# program to compute sum of
// all element after doing XOR
// with itself ( element_time)
using System;
 
class GFG {
     
    // function return sum of all
    // XOR element of array
    static int XorSum(int []arr, int n)
    {
         
        int result = 0;
        for (int i = 0; i < n; i++) {
     
        // if number is odd then add it to the
        // result else not
        if (arr[i] % 2 != 0)
            result += arr[i];
        }
     
        return result;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = {1, 2, 6, 3, 4, 5};
        int n = arr.Length;
        Console.WriteLine(XorSum(arr, n));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to compute
// sum of all element after
// doing XOR with itself
// ( element_time)
 
// function return sum of
// all XOR element of array
function XorSum($arr, $n)
{
    $result = 0;
    for ($i = 0; $i < $n; $i++)
    {
 
        // if number is odd
        // then add it to the
        // result else not
        if ($arr[$i] % 2 != 0)
            $result += $arr[$i];
    }
 
    return $result;
}
 
    // Driver Code
    $arr = array(1, 2, 6, 3, 4, 5);
    $n = count($arr);
    echo XorSum($arr, $n);
 
// This code is contributed by anuj_67.
?>

Javascript




<script>
 
// Javascript program to compute
// sum of all element after
// doing XOR with itself ( element_time)
 
// function return sum of all XOR element of array
function XorSum(arr, n)
{
    let result = 0;
    for (let i = 0; i < n; i++) {
 
        // if number is odd then add it to the
        // result else not
        if (arr[i] % 2 != 0)
            result += arr[i];
    }
 
    return result;
}
 
// Driver program
    let arr = [ 1, 2, 6, 3, 4, 5 ];
    let n = arr.length;
    document.write(XorSum(arr, n));
 
</script>

Output

9

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

This article is contributed by praveen kumar . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. 


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