Skip to content
Related Articles

Related Articles

Count pairs with set bits sum equal to K
  • Last Updated : 15 Jul, 2019

Given an array arr[] and an integer K, the task is to count the pairs whose sum of set bits is K

Examples:

Input: arr[] = {1, 2, 3, 4, 5}, K = 4
Output: 1
(3, 5) is the only valid pair as the count
of set bits in the integers {1, 2, 3, 4, 5}
are {1, 1, 2, 1, 2} respectively.

Input: arr[] = {5, 42, 35, 22, 7}, K = 6
Output: 6

Naive approach: Initialise count = 0 and run two nested loops and check all possible pairs and check whether the sum of count bits is K. If yes then increment count.



Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count
// of set bits in n
unsigned int countSetBits(int n)
{
    unsigned int count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
  
// Function to return the count
// of required pairs
int pairs(int arr[], int n, int k)
{
  
    // To store the count
    int count = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
  
            // Sum of set bits in both the integers
            int sum = countSetBits(arr[i])
                      + countSetBits(arr[j]);
  
            // If current pair satisfies
            // the given condition
            if (sum == k)
                count++;
        }
    }
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;
    cout << pairs(arr, n, k);
  
    return 0;
}


Java




// Java implementation of the approach
class GFG 
{
  
// Function to return the count
// of set bits in n
static int countSetBits(int n)
{
    int count = 0;
    while (n > 0
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
  
// Function to return the count
// of required pairs
static int pairs(int arr[], int n, int k)
{
  
    // To store the count
    int count = 0;
    for (int i = 0; i < n; i++) 
    {
        for (int j = i + 1; j < n; j++) 
        {
  
            // Sum of set bits in both the integers
            int sum = countSetBits(arr[i])
                    + countSetBits(arr[j]);
  
            // If current pair satisfies
            // the given condition
            if (sum == k)
                count++;
        }
    }
    return count;
}
  
// Driver code
public static void main(String args[]) 
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = arr.length;
    int k = 4;
    System.out.println(pairs(arr, n, k));
}
}
  
// This code has been contributed by 29AjayKumar


Python3




# Python3 implementation of the approach 
  
# Function to return the count 
# of set bits in n 
def countSetBits(n) :
  
    count = 0
    while (n) :
          
        n &= (n - 1); 
        count += 1
          
    return count;
  
  
# Function to return the count 
# of required pairs 
def pairs(arr, n, k) :
  
    # To store the count 
    count = 0
    for i in range(n) : 
        for j in range(i + 1, n) :
  
            # Sum of set bits in both the integers 
            sum = countSetBits(arr[i]) + countSetBits(arr[j]); 
  
            # If current pair satisfies 
            # the given condition 
            if (sum == k) :
                count += 1
                  
    return count; 
  
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 1, 2, 3, 4, 5 ]; 
      
    n = len(arr); 
    k = 4
      
    print(pairs(arr, n, k));
      
# This code is contributed by AnkitRai01


C#




// C# implementation of the approach 
using System;
public class GFG 
  
// Function to return the count 
// of set bits in n 
static int countSetBits(int n) 
    int count = 0; 
    while (n > 0) 
    
        n &= (n - 1); 
        count++; 
    
    return count; 
  
// Function to return the count 
// of required pairs 
static int pairs(int []arr, int n, int k) 
  
    // To store the count 
    int count = 0; 
    for (int i = 0; i < n; i++) 
    
        for (int j = i + 1; j < n; j++) 
        
  
            // Sum of set bits in both the integers 
            int sum = countSetBits(arr[i]) 
                    + countSetBits(arr[j]); 
  
            // If current pair satisfies 
            // the given condition 
            if (sum == k) 
                count++; 
        
    
    return count; 
  
// Driver code 
public static void Main(String []args) 
    int []arr = { 1, 2, 3, 4, 5 }; 
    int n = arr.Length; 
    int k = 4; 
    Console.WriteLine(pairs(arr, n, k)); 
  
// This code is contributed by Princi Singh


Output:

1

Time complexity: O(n2)

Efficient approach: Assume that every integer can be represented using 32 bits, create a frequency array freq[] of size 32 where freq[i] will store the count of numbers having set bits equal to i. Now run two nested loops on this frequency array, if i + j = K then count of pairs will be freq[i] * freq[j] for all such i and j.

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 32
  
// Function to return the count
// of set bits in n
unsigned int countSetBits(int n)
{
    unsigned int count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
  
// Function to return the count
// of required pairs
int pairs(int arr[], int n, int k)
{
  
    // To store the count
    int count = 0;
  
    // Frequency array
    int f[MAX + 1] = { 0 };
    for (int i = 0; i < n; i++)
        f[countSetBits(arr[i])]++;
  
    for (int i = 0; i <= MAX; i++) {
        for (int j = i; j <= MAX; j++) {
  
            // If current pair satisfies
            // the given condition
            if (i + j == k) {
  
                // (arr[i], arr[i]) cannot be a valid pair
                if (i == j)
                    count += ((f[i] * (f[i] - 1)) / 2);
                else
                    count += (f[i] * f[j]);
            }
        }
    }
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;
    cout << pairs(arr, n, k);
  
    return 0;
}


Java




// Java implementation of the approach
import java.util.*;
  
class GFG 
{
      
static int MAX = 32;
  
// Function to return the count
// of set bits in n
static int countSetBits(int n)
{
    int count = 0;
    while (n > 0
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
  
// Function to return the count
// of required pairs
static int pairs(int arr[], int n, int k)
{
  
    // To store the count
    int count = 0;
  
    // Frequency array
    int []f = new int[MAX + 1];
    for (int i = 0; i < n; i++)
        f[countSetBits(arr[i])]++;
  
    for (int i = 0; i <= MAX; i++)
    {
        for (int j = i; j <= MAX; j++)
        {
  
            // If current pair satisfies
            // the given condition
            if (i + j == k)
            {
  
                // (arr[i], arr[i]) cannot be a valid pair
                if (i == j)
                    count += ((f[i] * (f[i] - 1)) / 2);
                else
                    count += (f[i] * f[j]);
            }
        }
    }
    return count;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = arr.length;
    int k = 4;
    System.out.println(pairs(arr, n, k));
}
}
  
// This code is contributed by Rajput-Ji


Python3




# Python implementation of the approach 
MAX = 32
  
# Function to return the count 
# of set bits in n 
def countSetBits(n) :
    count = 0
    while (n): 
        n &= (n - 1); 
        count += 1
  
    return count; 
  
# Function to return the count 
# of required pairs 
def pairs(arr, n, k):
  
    # To store the count 
    count = 0
  
    # Frequency array 
    f = [0 for i in range(MAX + 1)]
  
    for i in range(n): 
        f[countSetBits(arr[i])] += 1
  
    for i in range(MAX + 1):
        for j in range(1, MAX + 1):
  
            # If current pair satisfies 
            # the given condition 
            if (i + j == k): 
  
                # (arr[i], arr[i]) cannot be a valid pair 
                if (i == j):
                    count += ((f[i] * (f[i] - 1)) / 2); 
                else:
                    count += (f[i] * f[j]); 
      
    return count; 
  
# Driver code 
arr = [ 1, 2, 3, 4, 5 ]
n = len(arr)
k = 4
  
print (pairs(arr, n, k)) 
  
# This code is contributed by CrazyPro


C#




// C# implementation of the approach
using System;
      
class GFG 
{
      
static int MAX = 32;
  
// Function to return the count
// of set bits in n
static int countSetBits(int n)
{
    int count = 0;
    while (n > 0) 
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
  
// Function to return the count
// of required pairs
static int pairs(int []arr, int n, int k)
{
  
    // To store the count
    int count = 0;
  
    // Frequency array
    int []f = new int[MAX + 1];
    for (int i = 0; i < n; i++)
        f[countSetBits(arr[i])]++;
  
    for (int i = 0; i <= MAX; i++)
    {
        for (int j = i; j <= MAX; j++)
        {
  
            // If current pair satisfies
            // the given condition
            if (i + j == k)
            {
  
                // (arr[i], arr[i]) cannot be a valid pair
                if (i == j)
                    count += ((f[i] * (f[i] - 1)) / 2);
                else
                    count += (f[i] * f[j]);
            }
        }
    }
    return count;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;
    int k = 4;
    Console.WriteLine(pairs(arr, n, k));
}
}
/* This code is contributed by PrinciRaj1992 */


Output:

1

Time complexity: O(n)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :