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Count pairs with set bits sum equal to K

  • Last Updated : 08 Apr, 2021

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:
(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:
 

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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

Javascript




<script>
 
// javascript implementation of the approach
 
// Function to return the count
// of set bits in n
function countSetBits(n)
{
    var count = 0;
    while (n > 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count
// of required pairs
function pairs(arr , n , k)
{
 
    // To store the count
    var count = 0;
    for (i = 0; i < n; i++)
    {
        for (j = i + 1; j < n; j++)
        {
 
            // Sum of set bits in both the integers
            var sum = countSetBits(arr[i])
                    + countSetBits(arr[j]);
 
            // If current pair satisfies
            // the given condition
            if (sum == k)
                count++;
        }
    }
    return count;
}
 
// Driver code
 
var arr = [ 1, 2, 3, 4, 5 ];
var n = arr.length;
var k = 4;
document.write(pairs(arr, n, k));
 
 
// This code contributed by shikhasingrajput
 
</script>
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 */

Javascript




<script>
 
// javascript implementation of the approach
var MAX = 32;
 
// Function to return the count
// of set bits in n
function countSetBits(n)
{
    var count = 0;
    while (n > 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count
// of required pairs
function pairs(arr , n , k)
{
 
    // To store the count
    var count = 0;
 
    // Frequency array
    var f = Array.from({length: MAX + 1}, (_, i) => 0);
    for (i = 0; i < n; i++)
        f[countSetBits(arr[i])]++;
 
    for (i = 0; i <= MAX; i++)
    {
        for (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
var arr = [ 1, 2, 3, 4, 5 ];
var n = arr.length;
var k = 4;
document.write(pairs(arr, n, k));
 
// This code contributed by Princi Singh
</script>
Output: 
1

 

Time complexity: O(n)
 




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