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# 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 Practice

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<``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

## 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

## Javascript

 ``

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 ``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;``    ``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 ` `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;``    ``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

 ``

## Javascript

 ``

Output

`8`

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