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Count of pairs in an Array with same number of set bits
  • Difficulty Level : Hard
  • Last Updated : 15 Dec, 2020

Given an array arr containing N integers, the task is to count the possible number of pairs of elements with the same number of set bits.

Examples: 

Input: N = 8, arr[] = {1, 2, 3, 4, 5, 6, 7, 8} 
Output:
Explanation: 
Elements with 1 set bit: 1, 2, 4, 8 
Elements with 2 set bits: 3, 5, 6 
Elements with 3 set bits: 7 
Hence, {1, 2}, {1, 4}, {1, 8}, {2, 4}, {2, 8}, {4, 8}, {3, 5}, {3, 6}, and {5, 6} are the possible such pairs.

Input: N = 12, arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} 
Output: 22 

Approach:  



  • Precompute and store the set bits for all numbers up to maximum element of the array in bitscount[]. For all powers of 2, store 1 at their respective index. After that, compute the set bits count for the remaining elements by the relation:

bitscount[i] = bitscount[previous power of 2] + bitscount[i – previous power of 2] 
 

  • Store the frequency of set bits in the array elements in a Map.
  • Add the number of possible pairs for every set bit count. If X elements have same number of set bits, the number of possible pairs among them is X * (X – 1) / 2.
  • Print the total count of such pairs.

Below code is the implementation of the above approach:

C++14

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// C++ Program to count
// possible number of pairs
// of elements with same
// number of set bits.
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the
// count of Pairs
int countPairs(int arr[], int N)
{
    // Get the maximum element
    int maxm = *max_element(arr, arr + N);
 
    int i, k;
    // Array to store count of bits
    // of all elements upto maxm
    int bitscount[maxm + 1] = { 0 };
 
    // Store the set bits
    // for powers of 2
    for (i = 1; i <= maxm; i *= 2)
        bitscount[i] = 1;
    // Compute the set bits for
    // the remaining elements
    for (i = 1; i <= maxm; i++) {
        if (bitscount[i] == 1)
            k = i;
        if (bitscount[i] == 0) {
            bitscount[i]
                = bitscount[k]
                  + bitscount[i - k];
        }
    }
 
    // Store the frequency
    // of respective counts
    // of set bits
    map<int, int> setbits;
    for (int i = 0; i < N; i++) {
        setbits[bitscount[arr[i]]]++;
    }
 
    int ans = 0;
    for (auto it : setbits) {
        ans += it.second
               * (it.second - 1) / 2;
    }
 
    return ans;
}
 
int main()
{
    int N = 12;
    int arr[] = { 1, 2, 3, 4, 5, 6, 7,
                  8, 9, 10, 11, 12 };
 
    cout << countPairs(arr, N);
 
    return 0;
}

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Java

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// Java program to count possible
// number of pairs of elements
// with same number of set bits
import java.util.*;
import java.lang.*;
 
class GFG{
     
// Function to return the
// count of Pairs
static int countPairs(int []arr, int N)
{
     
    // Get the maximum element
    int maxm = arr[0];
       
    for(int j = 1; j < N; j++)
    {
        if (maxm < arr[j])
        {
            maxm = arr[j];
        }
    }
   
    int i, k = 0;
       
    // Array to store count of bits
    // of all elements upto maxm
    int[] bitscount = new int[maxm + 1];
    Arrays.fill(bitscount, 0);
   
    // Store the set bits
    // for powers of 2
    for(i = 1; i <= maxm; i *= 2)
        bitscount[i] = 1;
           
    // Compute the set bits for
    // the remaining elements
    for(i = 1; i <= maxm; i++)
    {
        if (bitscount[i] == 1)
            k = i;
        if (bitscount[i] == 0)
        {
            bitscount[i] = bitscount[k] + 
                           bitscount[i - k];
        }
    }
   
    // Store the frequency
    // of respective counts
    // of set bits
    Map<Integer, Integer> setbits = new HashMap<>();
       
    for(int j = 0; j < N; j++) 
    {
        setbits.put(bitscount[arr[j]],
        setbits.getOrDefault(
            bitscount[arr[j]], 0) + 1);
    }
   
    int ans = 0;
   
    for(int it : setbits.values())
    {
        ans += it * (it - 1) / 2;
       
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 12;
    int []arr = { 1, 2, 3, 4, 5, 6, 7,
                  8, 9, 10, 11, 12 };
     
    System.out.println(countPairs(arr, N));
}
}
 
// This code is contributed by offbeat

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Python3

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# Python3 program to count possible number
# of pairs of elements with same number
# of set bits.
  
# Function to return the
# count of Pairs
def countPairs(arr, N):
     
    # Get the maximum element
    maxm = max(arr)
    i = 0
    k = 0
     
    # Array to store count of bits
    # of all elements upto maxm
    bitscount = [0 for i in range(maxm + 1)]
     
    i = 1
     
    # Store the set bits
    # for powers of 2
    while i <= maxm:
        bitscount[i] = 1
        i *= 2
         
    # Compute the set bits for
    # the remaining elements
    for i in range(1, maxm + 1):
        if (bitscount[i] == 1):
            k = i
        if (bitscount[i] == 0):
            bitscount[i] = (bitscount[k] +
                            bitscount[i - k])
  
    # Store the frequency
    # of respective counts
    # of set bits
    setbits = dict()
     
    for i in range(N):
        if bitscount[arr[i]] in setbits:
            setbits[bitscount[arr[i]]] += 1
        else:
            setbits[bitscount[arr[i]]] = 1
  
    ans = 0
     
    for it in setbits.values():
        ans += it * (it - 1) // 2
  
    return ans
  
# Driver Code
if __name__=='__main__':
     
    N = 12
    arr = [ 1, 2, 3, 4, 5, 6, 7,
            8, 9, 10, 11, 12 ]
  
    print(countPairs(arr, N))
  
# This code is contributed by pratham76

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C#

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// C# program to count
// possible number of pairs
// of elements with same
// number of set bits.
using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;
 
class GFG{
     
// Function to return the
// count of Pairs
static int countPairs(int []arr, int N)
{
     
    // Get the maximum element
    int maxm = -int.MaxValue;
     
    for(int j = 0; j < N; j++)
    {
        if (maxm < arr[j])
        {
            maxm = arr[j];
        }
    }
 
    int i, k = 0;
     
    // Array to store count of bits
    // of all elements upto maxm
    int []bitscount = new int[maxm + 1];
    Array.Fill(bitscount, 0);
 
    // Store the set bits
    // for powers of 2
    for(i = 1; i <= maxm; i *= 2)
        bitscount[i] = 1;
         
    // Compute the set bits for
    // the remaining elements
    for(i = 1; i <= maxm; i++)
    {
        if (bitscount[i] == 1)
            k = i;
        if (bitscount[i] == 0)
        {
            bitscount[i] = bitscount[k] +
                           bitscount[i - k];
        }
    }
 
    // Store the frequency
    // of respective counts
    // of set bits
    Dictionary<int,
               int> setbits = new Dictionary<int,
                                             int>();
     
    for(int j = 0; j < N; j++)
    {
        if (setbits.ContainsKey(bitscount[arr[j]]))
        {
            setbits[bitscount[arr[j]]]++;
        }
        else
        {
            setbits[bitscount[arr[j]]] = 1;
        }
    }
 
    int ans = 0;
 
    foreach(KeyValuePair<int, int> it in setbits)
    {
        ans += it.Value * (it.Value - 1) / 2;
    }
    return ans;
}
     
// Driver Code
public static void Main(string[] args)
{
    int N = 12;
    int []arr = { 1, 2, 3, 4, 5, 6, 7,
                  8, 9, 10, 11, 12 };
 
    Console.Write(countPairs(arr, N));
}
}
 
// This code is contributed by rutvik_56

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Output: 

22

 

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