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

Given an array of integers, the task is to compute sum of all array element 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. 
?> 

Output:

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. 
?> 

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 contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

PHP

C++

Java

Python3

C#

PHP

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