Sum of bit differences among all pairs

Given an integer array of n integers, find sum of bit differences in all pairs that can be formed from array elements. Bit difference of a pair (x, y) is count of different bits at same positions in binary representations of x and y. 

For example, bit difference for 2 and 7 is 2. Binary representation of 2 is 010 and 7 is 111 ( first and last bits differ in two numbers). 

Examples :

Input: arr[] = {1, 2}
Output: 4
All pairs in array are (1, 1), (1, 2)
                       (2, 1), (2, 2)
Sum of bit differences = 0 + 2 +
                         2 + 0
                      = 4

Input:  arr[] = {1, 3, 5}
Output: 8
All pairs in array are (1, 1), (1, 3), (1, 5)
                       (3, 1), (3, 3) (3, 5),
                       (5, 1), (5, 3), (5, 5)
Sum of bit differences =  0 + 1 + 1 +
                          1 + 0 + 2 +
                          1 + 2 + 0 
                       = 8

Source: Google Interview Question

Naive Solution:

A Simple Solution is to run two loops to consider all pairs one by one. For every pair, count bit differences. Finally return sum of counts. Time complexity of this solution is O(n²). We are using bitset::count() which is an inbuilt STL in C++ which returns the number of set bits in the binary representation of a number. 

C++

// C++ program to compute sum of pairwise bit differences
#include <bits/stdc++.h>
using namespace std;

int sum_bit_diff(vector<int> a)
{
    int n = a.size();
    int ans = 0;

    for (int i = 0; i < n - 1; i++) {
        int count = 0;

        for (int j = i; j < n; j++) {
            // Bitwise and of pair (a[i], a[j])
            int x = a[i] & a[j];
            // Bitwise or of pair (a[i], a[j])
            int y = a[i] | a[j];

            bitset<32> b1(x);
            bitset<32> b2(y);

            // to count set bits in and of two numbers
            int r1 = b1.count();
            // to count set bits in or of two numbers
            int r2 = b2.count();

            // Absolute differences at individual bit positions of two
            // numbers is contributed by pair (a[i], a[j]) in count
            count = abs(r1 - r2);

            // each pair adds twice of contributed count
            // as both (a, b) and (b, a) are considered
            // two separate pairs.
            ans = ans + (2 * count);
        }
    }
    return ans;
}

int main()
{

    vector<int> nums{ 10, 5 };
    int ans = sum_bit_diff(nums);

    cout << ans;
}

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {

    static int sumBitDiff(int[] arr){
        int diff = 0;                                //hold the ans

          for(int i=0; i<arr.length; i++){
            for(int j=i; j<arr.length; j++){

              //XOR toggles the bits and will form a number that will have
              //set bits at the places where the numbers bits differ
              //eg: 010 ^ 111 = 101...diff of bits = count of 1's = 2

                 int xor = arr[i]^arr[j];
                  int count = countSetBits(xor);        //Integer.bitCount() can also be used

                  //when i == j (same numbers) the xor would be 0,
                  //thus our ans will remain unaffected as (2*0 = 0)
                  diff += 2*count;
            }
        }

          return diff;
    }

    //Kernighan algo
      static int countSetBits(int n){
        int count = 0;            // `count` stores the total bits set in `n`

        while (n != 0) {
            n = n & (n - 1);    // clear the least significant bit set
            count++;
        }

        return count;
    }

    public static void main (String[] args) {
        int[] arr = {5,10};
          int ans  = sumBitDiff(arr);
        System.out.println(ans);
    }
}

Python3

# Python3 program for the above approach
def sumBitDiff(arr):
    diff = 0    #hold the ans

    for i in range(len(arr)):
        for j in range(i, len(arr)):

        # XOR toggles the bits and will form a number that will have
        # set bits at the places where the numbers bits differ
        # eg: 010 ^ 111 = 101...diff of bits = count of 1's = 2
            xor = arr[i]^arr[j]
            count = countSetBits(xor)        #Integer.bitCount() can also be used

            # when i == j (same numbers) the xor would be 0,
            # thus our ans will remain unaffected as (2*0 = 0)
            diff += (2*count)

    return diff

# Kernighan algo
def countSetBits(n):
    count = 0            # `count` stores the total bits set in `n`

    while (n != 0) :
        n = n & (n - 1)    # clear the least significant bit set
        count += 1

    return count

# Driver code
if __name__ == "__main__":

    arr = [5,10]
    ans  = sumBitDiff(arr)
    print(ans)

    # This code is contributed by sanjoy_62.

C#

/*package whatever //do not write package name here */

using System;

public class GFG {

    static int sumBitDiff(int[] arr){
        int diff = 0;                                //hold the ans

          for(int i=0; i<arr.Length; i++){
            for(int j=i; j<arr.Length; j++){

              //XOR toggles the bits and will form a number that will have
              //set bits at the places where the numbers bits differ
              //eg: 010 ^ 111 = 101...diff of bits = count of 1's = 2

                 int xor = arr[i]^arr[j];
                  int count = countSetBits(xor);        //int.bitCount() can also be used

                  //when i == j (same numbers) the xor would be 0,
                  //thus our ans will remain unaffected as (2*0 = 0)
                  diff += 2*count;
            }
        }

          return diff;
    }

    //Kernighan algo
      static int countSetBits(int n){
        int count = 0;            // `count` stores the total bits set in `n`

        while (n != 0) {
            n = n & (n - 1);    // clear the least significant bit set
            count++;
        }

        return count;
    }

    public static void Main(String[] args) {
        int[] arr = {5,10};
          int ans  = sumBitDiff(arr);
        Console.WriteLine(ans);
    }
}

// This code contributed by umadevi9616

Javascript

<script>

// javascript program for above approach
    function sumBitDiff(arr)
    {
        let diff = 0;
        // hold the ans

          for(let i = 0; i < arr.length; i++){
            for(let j = i; j < arr.length; j++){

              // XOR toggles the bits and will form a number that will have
              // set bits at the places where the numbers bits differ
              // eg: 010 ^ 111 = 101...diff of bits = count of 1's = 2

                 let xor = arr[i]^arr[j];
                  let count = countSetBits(xor);
                  // Integer.bitCount() can also be used

                  // when i == j (same numbers) the xor would be 0,
                  // thus our ans will remain unaffected as (2*0 = 0)
                  diff += 2*count;
            }
        }

          return diff;
    }

    // Kernighan algo
      function countSetBits(n){
        let count = 0;            // `count` stores the total bits set in `n`

        while (n != 0) {
            n = n & (n - 1);    // clear the least significant bit set
            count++;
        }

        return count;
    }


// Driver code
    let arr = [5,10];
    let ans  = sumBitDiff(arr);
    document.write(ans);

    // This code is contributed by splevel62.
</script>

Output

8

Time Complexity: O(n²)
Auxiliary Space: O(1)

Efficient Solution :

An Efficient Solution can solve this problem in O(n) time using the fact that all numbers are represented using 32 bits (or some fixed number of bits). The idea is to count differences at individual bit positions. We traverse from 0 to 31 and count numbers with i’th bit set. Let this count be ‘count’. There would be “n-count” numbers with i’th bit not set. So count of differences at i’th bit would be “count * (n-count) * 2”, the reason for this formula is as every pair having one element which has set bit at i’th position and second element having unset bit at i’th position contributes exactly 1 to sum, therefore total permutation count will be count*(n-count) and multiply by 2 is due to one more repetition of all this type of pair as per given condition for making pair 1<=i, j<=N.

Below is implementation of above idea.  

C++

// C++ program to compute sum of pairwise bit differences
#include <bits/stdc++.h>
using namespace std;

int sumBitDifferences(int arr[], int n)
{
    int ans = 0; // Initialize result
    // traverse over all bits
    for (int i = 0; i < 32; i++) {
        // count number of elements with i'th bit set
        int count = 0;
        for (int j = 0; j < n; j++)
            if ((arr[j] & (1 << i)))
                count++;
        // Add "count * (n - count) * 2" to the answer
        ans += (count * (n - count) * 2);
    }
    return ans;
}

// Driver program
int main()
{
    int arr[] = { 1, 3, 5 };
    int n = sizeof arr / sizeof arr[0];
    cout << sumBitDifferences(arr, n) << endl;
    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

C

// C program to compute sum of pairwise bit differences
#include <stdio.h>

int sumBitDifferences(int arr[], int n)
{
    int ans = 0; // Initialize result
    // traverse over all bits
    for (int i = 0; i < 32; i++) {
        // count number of elements with i'th bit set
        int count = 0;
        for (int j = 0; j < n; j++)
            if ((arr[j] & (1 << i)))
                count++;
        // Add "count * (n - count) * 2" to the answer
        ans += (count * (n - count) * 2);
    }
    return ans;
}

// Driver program
int main()
{
    int arr[] = { 1, 3, 5 };
    int n = sizeof arr / sizeof arr[0];
    printf("%d\n", sumBitDifferences(arr, n));
    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

Java

// Java program to compute sum of pairwise
// bit differences

import java.io.*;
class GFG {
    static int sumBitDifferences(int arr[], int n)
    {
        int ans = 0; // Initialize result
        // traverse over all bits
        for (int i = 0; i < 32; i++) {
            // count number of elements with i'th bit set
            int count = 0;
            for (int j = 0; j < n; j++)
                if ((arr[j] & (1 << i)) != 0)
                    count++;
            // Add "count * (n - count) * 2"
            // to the answer...(n - count = unset bit count)
            ans += (count * (n - count) * 2);
        }
        return ans;
    }

    // Driver program
    public static void main(String args[])
    {
        int arr[] = { 1, 3, 5 };
        int n = arr.length;
        System.out.println(sumBitDifferences(arr, n));
    }
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

Python3

# Python program to compute sum of pairwise bit differences

def sumBitDifferences(arr, n):

    ans = 0  # Initialize result

    # traverse over all bits
    for i in range(0, 32):

        # count number of elements with i'th bit set
        count = 0
        for j in range(0, n):
            if ( (arr[j] & (1 << i)) ):
                count+= 1

        # Add "count * (n - count) * 2" to the answer
        ans += (count * (n - count) * 2);

    return ans

# Driver program
arr = [1, 3, 5]
n = len(arr )
print(sumBitDifferences(arr, n))

# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to compute sum
// of pairwise bit differences
using System;

class GFG {
    static int sumBitDifferences(int[] arr,
                                 int n)
    {
        int ans = 0; // Initialize result

        // traverse over all bits
        for (int i = 0; i < 32; i++) {

            // count number of elements
            // with i'th bit set
            int count = 0;
            for (int j = 0; j < n; j++)
                if ((arr[j] & (1 << i)) != 0)
                    count++;

            // Add "count * (n - count) * 2"
            // to the answer
            ans += (count * (n - count) * 2);
        }

        return ans;
    }
    // Driver Code
    public static void Main()
    {

        int[] arr = { 1, 3, 5 };
        int n = arr.Length;

        Console.Write(sumBitDifferences(arr, n));
    }
}

// This code is contributed by ajit

PHP

<?php
// PHP program to compute sum
// of pairwise bit differences

function sumBitDifferences($arr, $n)
{
    // Initialize result
    $ans = 0;

    // traverse over all bits
    for ($i = 0; $i < 32; $i++)
    {
        // count number of elements
        // with i'th bit set
        $count = 0;
        for ($j = 0; $j < $n; $j++)
            if (($arr[$j] & (1 << $i)))
                $count++;

        // Add "count * (n - count) * 2"
        // to the answer
        $ans += ($count * ($n -
                           $count) * 2);
    }

    return $ans;
}

// Driver Code
$arr = array(1, 3, 5);
$n = sizeof($arr);
echo sumBitDifferences($arr, $n), "\n";

// This code is contributed by m_kit
?>

Javascript

<script>

// Javascript program to compute sum
// of pairwise bit differences
function sumBitDifferences(arr, n)
{

    // Initialize result
    let ans = 0;

    // Traverse over all bits
    for(let i = 0; i < 32; i++)
    {

        // count number of elements with i'th bit set
        let count = 0;
        for(let j = 0; j < n; j++)
            if ((arr[j] & (1 << i)))
                count++;

        // Add "count * (n - count) * 2" to the answer
        ans += (count * (n - count) * 2);
    }
    return ans;
}

// Driver code
let arr = [ 1, 3, 5 ];
let n = arr.length;

document.write(sumBitDifferences(arr, n));

// This code is contributed by subhammahato348

</script>

Output

8

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

Asked in: Google

Thanks to Gaurav Ahirwar for suggesting this solution.