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} 

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 <bits/stdc++.h>
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[0]);

    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<Integer, Integer> 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<Integer, Integer> 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

<script>

// Javascript implementation of above approach

 /* Function to get no of set
bits in binary representation
of passed binary no. */
function countSetBits(n)
{
    var count = 0;
    while (n > 0)
    {
        n &= (n - 1) ;
        count++;
    }
    return count;
}

// Function to count all pairs
// with equal set bits count
function totalPairs(arr, n)
{
    // map to store count of elements
    // with equal number of set bits
    var m = new Map();
    for (var i = 0; i < n; i++) {

        // inbuilt function that returns the
        // count of set bits of the number
        if(m.has(arr[i]))
        {
            m.set(countSetBits(arr[i]), m.get(countSetBits(arr[i]))+1);
        }
        else{
            m.set(countSetBits(arr[i]), 1);
        }

    }

    var result = 0;
    m.forEach((value, key) => {
        // there can be (n*(n-1)/2) unique two-
        // element pairs to choose from n elements
        result += parseInt(key * (key - 1) / 2);
    });

    return result;
}

// Driver code
var arr = [7, 5, 3, 9, 1, 2 ];
var n = arr.length;
document.write( totalPairs(arr, n));

</script>

Output:

4