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 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::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 mp = new Dictionary();         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 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

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