Sum of XOR of all pairs in an array

Given an array of n integers, find the sum of XOR of all pairs of numbers in the array.

Examples:

Input: arr[] = [7, 3, 5]
Output: 12
Explanation:
All possible pairs and there XOR
Value: (7 ^ 3 = 4) + (3 ^ 5 = 6)
(7 ^ 5 = 2) = 4 + 6 + 2 = 12

Input: arr[] = [5, 9, 7, 6]
Output: 47
Explanation:
All possible pairs and there XOR
Value: (5 ^ 9 = 12) + (5 ^ 7 = 2)
+ (5 ^ 6 = 3) + (9 ^ 7 = 14)
+ (9 ^ 6 = 15) + (7 ^ 6 = 1)
= 12 + 2 + 3 + 14 + 15 + 1 = 47

[Naive Approach] Iterative Method - O(n^2) Time and O(1) Space

Iterate over all unique pairs of elements and compute their XOR, adding each result to the total sum.

Dry run for arr[] = [7, 3, 5]:

• Initial sum = 0
• i = 0 (7)
  j = 1: 7 ^ 3 = 4 then add to sum, now sum = 4
  j = 2: 7 ^ 5 = 2 then add to sum, now sum = 6
• i = 1 (3)
  j = 2: 3 ^ 5 = 6 then add to sum, now sum = 12
• All unique pairs are processed, Final sum = 12

#include <iostream>
#include <vector>
using namespace std;

long sumXOR(vector<int> arr)
{
    long sum = 0;

    // Consider all pairs (arr[i], arr[j) such that
    // i < j
    for (int i = 0; i < arr.size() - 1; i++)
        for (int j = i + 1; j < arr.size(); j++)
            sum += arr[i] ^ arr[j];

    return sum;
}

int main()
{
    vector<int> arr = { 5, 9, 7, 6 };

    cout << sumXOR(arr);
    return 0;
}

class GFG {
    
    static int sumXOR(int arr[])
    {
        int sum = 0; 
    
        // Consider all pairs (arr[i], arr[j) 
        // such that i < j
        for (int i = 0; i < arr.length - 1; i++)
            for (int j = i + 1; j < arr.length; j++)
                sum += arr[i] ^ arr[j];
    
        return sum;
    }

    public static void main (String[] args) {
        int arr[] = { 5, 9, 7, 6 };

        System.out.println(sumXOR(arr));
    }
}

def sumXOR(arr) :
    sum = 0

    # Consider all pairs (arr[i], arr[j) 
    # such that i < j
    for i in range(0, len(arr)-1) :
        
        for j in range(i + 1, len(arr)) :
            
            sum = sum + (arr[i] ^ arr[j])
            
    return sum
    
if __name__ == "__main__":
    arr = [ 5, 9, 7, 6 ]
    print(sumXOR(arr, n))

using System;

class GFG {
    
    static int sumXOR(int []arr)
    {
        int sum = 0; 
    
        // Consider all pairs (arr[i], arr[j) 
        // such that i < j
        for (int i = 0; i < arr.Length-1; i++)
            for (int j = i + 1; j < arr.Length; j++)
                sum += arr[i] ^ arr[j];
    
        return sum;
    }

    public static void Main () {
        int []arr = { 5, 9, 7, 6 };
        
        Console.WriteLine(sumXOR(arr));
    }
}

const sumXOR = (arr, n) => {
    let sum = 0;

    // Consider all pairs (arr[i], arr[j) such that
    // i < j
    for (let i = 0; i < arr.length-1; i++)
        for (let j = i + 1; j < arr.length; j++)
            sum += arr[i] ^ arr[j];

        return sum;
}

// Driver code
let arr = [5, 9, 7, 6];
console.log(sumXOR(arr));

47

[Expected Approach] Bit Manipulation - O(n) Time and O(1) Space

Assume integers are represented using 32 bits. For each bit position, count how many numbers have 0 and how many have 1, then multiply these counts with 2^i (where i is the bit position) to get the contribution of that bit. Summing contributions from all bit positions gives the final answer.

How this actually works?

Let us consider rightmost bit as an example.. Suppose a numbers have 0-bit and b numbers have 1-bit at this position. Then, out of all pairs, a * b pairs will have 1 in this bit of the XOR, because there are a * b ways to choose one number with 0 and one with 1. So, this bit contributes a * b to the total XOR sum. In general, for the i-th bit (starting from 0), count numbers with 0 as ai and numbers with 1 as bi so, the contribution is ai * bi * 2^i. Repeat this for all bit positions and sum all contributions to get the final sum.

For arr[] = [7, 3, 5]

1. Binary representation:
   7 = 1 1 1
   3 = 0 1 1
   5 = 1 0 1
2. For bit position 0 (rightmost bit):
   Bits with zero = 0
   Bits with one = 3
   Contribution = 0 * 3 * 2^0 = 0
3. For bit position 1:
   Bits with zero = 1
   Bits with one = 2
   Contribution = 1 * 2 * 2^1 = 4
4. For bit position 2:
   Bits with zero = 1
   Bits with one = 2
   Contribution = 1 * 2 * 2^2 = 8
5. Final sum = 0 + 4 + 8 = 12

#include <iostream>
#include <vector>
using namespace std;

long sumXOR(vector<int>& arr)
{
    long sum = 0;
    for (int i = 0; i < 32; i++) 
    {
        //  Count of zeros and ones
        int zc = 0, oc = 0; 
        
        for (int j = 0; j < arr.size(); j++)
        {
            if ((arr[j] >> i) & 1)
                    oc++;
                else
                    zc++;
        }
        
        // Adding individual bit sum 
        sum += (long) oc * zc * (1 << i); 
    }
    return sum;
}

int main()
{
    vector<int> arr = { 5, 9, 7, 6 };

    cout << sumXOR(arr);
    return 0;
}

class GfG {
    
    static long sumXOR(int arr[])
    {
        long sum = 0;
        for (int i = 0; i < 32; i++) 
        {
            // Count of zeros and ones
            int zc = 0, oc = 0; 
            
            for (int j = 0; j < arr.length; j++)
            {
                if ((arr[j] >> i & 1) == 1)
                    oc++;
                else
                    zc++;
            }
            
            // Adding individual bit sum 
            sum += (long) oc * zc * (1 << i); 
        }
        return sum;
    }
    
    public static void main(String args[])
    {
        int arr[] = { 5, 9, 7, 6 };

        System.out.println(sumXOR(arr));
    }
}

def sumXOR(arr):
    
    sum = 0
    for i in range(0, 32):

        #  Count of zeros and ones
        zc = 0
        oc = 0
         
        for j in range(0, len(arr)):
            if (arr[j] >> i) & 1:
                oc = oc + 1
            else:
                zc = zc + 1
        
        # Adding individual bit sum 
        sum = sum + oc * zc * (1 << i)
    
    return sum

if __name__ == "__main__":
    arr = [  5, 9, 7, 6 ]
    print (sumXOR(arr))

using System;

class GfG {
    
    static long sumXOR(int []arr)
    {
        long sum = 0;
        for (int i = 0; i < 32; i++) 
        {
            // Count of zeros and ones
            int zc = 0, oc = 0; 
            
            for (int j = 0; j < arr.Length; j++)
            {
                if (((arr[j] >> i) & 1) == 1)
                    oc++;
                else
                    zc++;
            }
            
            // Adding individual bit sum
            sum += (long) oc * zc * (1 << i); 
        }
        return sum;
    }

    public static void Main()
    {
        int []arr = { 5, 9, 7, 6 };

        Console.WriteLine(sumXOR(arr));
    }
}

const sumXOR = (arr) => {
    let sum = 0;
    for (let i = 0; i < 32; i++) {

        // Count of zeros and ones
        let zc = 0, oc = 0;

        for (let j = 0; j < arr.length; j++) {
            if ((arr[j] >> i) & 1)
                oc++;
            else
                zc++;
        }

        // Adding individual bit sum
        sum += zc * oc * (1 << i);
    }
    return sum;
}

// Driver Code
let arr = [5, 9, 7, 6];

console.log(sumXOR(arr));

47