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Count pairs in an array such that both elements has equal set bits
• Difficulty Level : Basic
• Last Updated : 12 May, 2021

Given an array arr with unique elements, the task is to count the total number of pairs of elements that have equal set bits count.
Examples:

Input: arr[] = {2, 5, 8, 1, 3}
Output:
Set bits counts for {2, 5, 8, 1, 3} are {1, 2, 1, 1, 2}
All pairs with same set bits count are {2, 8}, {2, 1}, {5, 3}, {8, 1}
Input: arr[] = {1, 11, 7, 3}
Output:
Only possible pair is {11, 7}

Approach:

• Traverse the array from left to right and count total number of set bits of each integer.
• Use a map to store the number of elements with same count of set bits with set bits as key, and count as value.
• Then iterator through map elements, and calculate how many two element pairs can be formed from n elements (for each element of the map) i.e. (n * (n-1)) / 2.
• Final result will be the sum of output from the previous step for every element of the map.

Below is the implementation of the of the above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Function to count all pairs``// with equal set bits count``int` `totalPairs(``int` `arr[], ``int` `n)``{``    ``// map to store count of elements``    ``// with equal number of set bits``    ``map<``int``, ``int``> m;``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// inbuilt function that returns the``        ``// count of set bits of the number``        ``m[__builtin_popcount(arr[i])]++;``    ``}` `    ``map<``int``, ``int``>::iterator it;``    ``int` `result = 0;``    ``for` `(it = m.begin(); it != m.end(); it++) {` `        ``// there can be (n*(n-1)/2) unique two-``        ``// element pairs to choose from n elements``        ``result``            ``+= (*it).second * ((*it).second - 1) / 2;``    ``}` `    ``return` `result;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 7, 5, 3, 9, 1, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << totalPairs(arr, n);` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `class` `GFG {``    ` `    ``/* Function to get no of set ``    ``bits in binary representation ``    ``of passed binary no. */``    ``static` `int` `countSetBits(``int` `n)``    ``{``        ``int` `count = ``0``;``        ``while` `(n > ``0``)``        ``{``            ``n &= (n - ``1``) ;``            ``count++;``        ``}``        ``return` `count;``    ``}``    ` `    ``// Function to count all pairs``    ``// with equal set bits count``    ``static` `int` `totalPairs(``int` `arr[], ``int` `n)``    ``{``        ``// map to store count of elements``        ``// with equal number of set bits``        ``HashMap m = ``new` `HashMap<>();``        ``for` `(``int` `i = ``0``; i < n; i++) {``    ` `            ``// function that returns the``            ``// count of set bits of the number``            ``int` `count = countSetBits(arr[i]);``            ``if``(m.containsKey(count))``                ``m.put(count, m.get(count) + ``1``);``            ``else``                ``m.put(count, ``1``);``        ``}``    ` `        ``int` `result = ``0``;``        ``for` `(Map.Entry entry : m.entrySet()) {``            ``int` `value = entry.getValue();``            ` `            ``// there can be (n*(n-1)/2) unique two-``            ``// element pairs to choose from n elements``            ``result += ((value * (value -``1``)) / ``2``);``        ``}``    ` `        ``return` `result;``    ``}``    ` `    ``public` `static` `void` `main (String[] args) {``        ``int` `arr[] = { ``7``, ``5``, ``3``, ``9``, ``1``, ``2` `};``        ``int` `n = arr.length;``        ``System.out.println(totalPairs(arr, n));``    ``}``}`

## Python3

 `# Python3 implementation of the above approach` `# Function to count all pairs``# with equal set bits count``def` `totalPairs(arr, n):``    ` `    ``# map to store count of elements``    ``# with equal number of set bits``    ``m ``=` `dict``()` `    ``for` `i ``in` `range``(n):` `        ``# inbuilt function that returns the``        ``# count of set bits of the number``        ``x ``=` `bin``(arr[i]).count(``'1'``)` `        ``m[x] ``=` `m.get(x, ``0``) ``+` `1``;` `    ``result ``=` `0``    ``for` `it ``in` `m:` `        ``# there can be (n*(n-1)/2) unique two-``        ``# element pairs to choose from n elements``        ``result``+``=` `(m[it] ``*` `(m[it] ``-` `1``)) ``/``/` `2``    ` `    ``return` `result` `# Driver code``arr ``=` `[``7``, ``5``, ``3``, ``9``, ``1``, ``2``]``n ``=` `len``(arr)` `print``(totalPairs(arr, n))` `# This code is contributed by mohit kumar`

## C#

 `// C# program to rearrange a string so that all same``// characters become atleast d distance away``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ` `    ``/* Function to get no of set``    ``bits in binary representation``    ``of passed binary no. */``    ``static` `int` `countSetBits(``int` `n)``    ``{``        ``int` `count = 0;``        ``while` `(n > 0)``        ``{``            ``n &= (n - 1) ;``            ``count++;``        ``}``        ``return` `count;``    ``}``    ` `    ``// Function to count all pairs``    ``// with equal set bits count``    ``static` `int` `totalPairs(``int` `[]arr, ``int` `n)``    ``{``        ``// map to store count of elements``        ``// with equal number of set bits``        ``Dictionary<``int``,``int``> mp = ``new` `Dictionary<``int``,``int``>();``        ``for` `(``int` `i = 0 ; i < n; i++)``        ``{``            ``// function that returns the``            ``// count of set bits of the number``            ``int` `count = countSetBits(arr[i]);``            ``if``(mp.ContainsKey(count))``            ``{``                ``var` `val = mp[count];``                ``mp.Remove(count);``                ``mp.Add(count, val + 1);``            ``}``            ``else``            ``{``                ``mp.Add(count, 1);``            ``}``        ``}``    ` `        ``int` `result = 0;``        ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `mp){``            ``int` `values = entry.Value;``            ` `            ``// there can be (n*(n-1)/2) unique two-``            ``// element pairs to choose from n elements``            ``result += ((values * (values -1)) / 2);``        ``}``    ` `        ``return` `result;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main (String[] args)``    ``{``        ``int` `[]arr = { 7, 5, 3, 9, 1, 2 };``        ``int` `n = arr.Length;``        ``Console.WriteLine(totalPairs(arr, n));``    ``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``
Output:
`4`

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