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:

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// 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;
}

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Output:

4


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