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Sum of bit differences among all pairs

  • Difficulty Level : Hard
  • Last Updated : 17 Jul, 2021

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);
    }
}
Output



8

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

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 prorgram
    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 Anshika Goyal.

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 prorgram
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
 

 

Asked in: Google

Thanks to Gaurav Ahirwar for suggesting this solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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