# 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

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 using namespace std;    int sum_bit_diff(vector 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 nums{ 10, 5 };     int ans = sum_bit_diff(nums);        cout << ans; }

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 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             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



Output :

8

Thanks to Gaurav Ahirwar for suggesting this solution.

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