# Count pairs in an array such that both elements has equal set bits

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: 4
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: 1
Only possible pair is {11, 7}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

Output:

```4
```

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.