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Count of pairs {X, Y} from an array such that sum of count of set bits in X ⊕ Y and twice the count of set bits in X & Y is M

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Given an array arr[] consisting of N non-negative integers and an integer M, the task is to find the count of unordered pairs {X, Y} of array elements that satisfies the condition setBits(X ⊕ Y) + 2 * setBits(X & Y) = M, where denotes the Bitwise XOR and & denotes the Bitwise AND.

Examples:

Input: arr[] = {30, 0, 4, 5 }, M = 2
Output: 2
Explanation: The pairs satisfying the necessary conditions are {{3, 0}, {0, 5}}.

Input: arr[] = {1, 2, 3, 4}, M = 3
Output: 3
Explanation: The pairs satisfying the necessary conditions are {{1, 3}, {2, 3}, {3, 4}}.

 

Naive Approach: The simplest approach is to generate all possible pairs from the given array and for each pair, check if the necessary condition is satisfied or not. Increment the count for pairs satisfying the given conditions and finally, print the count of pairs.

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find total count of
// given pairs satisfying the equation
int countPairs(int a[], int N, int M)
{
 
    // Initialize the count to 0
    int count = 0;
 
    // Loop through the array
    for (int i = 0; i < N; i++) {
 
        // Loop through the remaining
        // elements of the array
        for (int j = i + 1; j < N; j++) {
 
            // Calculate the XOR of
            // the pair elements
            int x = a[i] ^ a[j];
 
            // Calculate the AND of
            // the pair elements
            int y = a[i] & a[j];
 
            // Check if the condition is satisfied
            // __builtin_popcount() returns no. of set bits
            if (__builtin_popcount(x)
                    + 2 * __builtin_popcount(y)
                == M) {
 
                // increment the count
                count++;
            }
        }
    }
 
    // Return the count of pairs
    // satisfying the condition
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 0, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int M = 2;
 
    // Function Call
    cout << countPairs(arr, N, M) << endl;
 
    return 0;
}

                    

Java

// Java code
 
import java.io.*;
import java.util.*;
 
public class CountPairs {
 
    // Function to find the total count of
    // given pairs satisfying the equation
    static int countPairs(int[] a, int N, int M) {
        // Initialize the count to 0
        int count = 0;
 
        // Loop through the array
        for (int i = 0; i < N; i++) {
            // Loop through the remaining
            // elements of the array
            for (int j = i + 1; j < N; j++) {
                // Calculate the XOR of the pair elements
                int x = a[i] ^ a[j];
 
                // Calculate the AND of the pair elements
                int y = a[i] & a[j];
 
                // Check if the condition is satisfied
                // Integer.bitCount() returns the number of set bits
                if (Integer.bitCount(x) + 2 * Integer.bitCount(y) == M) {
                    // Increment the count
                    count++;
                }
            }
        }
 
        // Return the count of pairs satisfying the condition
        return count;
    }
 
    // Driver code
    public static void main(String[] args) {
        int arr[] = { 3, 0, 4, 5 };
        int N = arr.length;
        int M = 2;
 
        // Function Call
        System.out.println(countPairs(arr, N, M));
    }
}
 
 
// This code is contributed by guptapratik

                    

Python3

# Python3 Program for the above approach
 
# Function to count number of setbits in the number n
def  countSetBits(n):
    # Stores the count of setbits
    count = 0   
    # Iterate while N is greater than 0
    while (n):  
        # Increment count by 1 if N is odd
        count += n & 1
         
        # Right shift N
        n >>= 1
         
    # Return the count of set bits
    return count
# Function to find total count of
# given pairs satisfying the equation
def countPairs(a, N, M):
    #Initialize the count to 0
    count = 0
 
    #Loop through the array
    for i in range(0,N):
 
        #Loop through the remaining
        #elements of the array
        for j in range(i+1,N):
            # Calculate the XOR of the pair elements
            x = a[i] ^ a[j]
         
            # Calculate the AND of the pair elements
            y = a[i] & a[j]
 
            if (countSetBits(x) + 2 * countSetBits(y))== M:
                #increment the count
                count+=1
 
    # Return the count of pairs satisfying the condition
    return count
     
         
arr = [ 3, 0, 4, 5]
N = len(arr)
M = 2
# Function call
pairs=countPairs(arr, N, M)
print(pairs)

                    

C#

using System;
 
namespace CountPairs {
class Program {
    // Function to find total count of
    // given pairs satisfying the equation
    static int CountPairs(int[] a, int N, int M)
    {
        // Initialize the count to 0
        int count = 0;
 
        // Loop through the array
        for (int i = 0; i < N; i++) {
            // Loop through the remaining
            // elements of the array
            for (int j = i + 1; j < N; j++) {
                // Calculate the XOR of
                // the pair elements
                int x = a[i] ^ a[j];
 
                // Calculate the AND of
                // the pair elements
                int y = a[i] & a[j];
 
                // Check if the condition is satisfied
                // CountSetBits() returns no. of set bits
                if (CountSetBits(x) + 2 * CountSetBits(y)
                    == M) {
                    // increment the count
                    count++;
                }
            }
        }
 
        // Return the count of pairs
        // satisfying the condition
        return count;
    }
 
    // Function to count the number of set bits
    static int CountSetBits(int num)
    {
        int count = 0;
        while (num > 0) {
            count += num & 1;
            num >>= 1;
        }
        return count;
    }
 
    // Driver code
    static void Main(string[] args)
    {
        int[] arr = { 3, 0, 4, 5 };
        int N = arr.Length;
        int M = 2;
 
        // Function Call
        Console.WriteLine(CountPairs(arr, N, M));
    }
}
}

                    

Javascript

// Function to find total count of
// given pairs satisfying the equation
function countPairs(arr, M) {
    // Initialize the count to 0
    let count = 0;
 
    // Loop through the array
    for (let i = 0; i < arr.length; i++) {
        // Loop through the remaining
        // elements of the array
        for (let j = i + 1; j < arr.length; j++) {
            // Calculate the XOR of the pair elements
            let x = arr[i] ^ arr[j];
 
            // Calculate the AND of the pair elements
            let y = arr[i] & arr[j];
 
            // Check if the condition is satisfied
            if (popcount(x) + 2 * popcount(y) === M) {
                // increment the count
                count++;
            }
        }
    }
 
    // Return the count of pairs satisfying the condition
    return count;
}
 
// Helper function to count the number of set bits in a number
function popcount(num) {
    let count = 0;
    while (num > 0) {
        count += num & 1;
        num >>= 1;
    }
    return count;
}
 
// Driver code
let arr = [3, 0, 4, 5];
let M = 2;
 
// Function Call
console.log(countPairs(arr, M));

                    


Output: 

2

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations: 

  • From the property of Bitwise XOR:
    • setBits( a⊕ b ) = setBits( a|b ) – setBits( a&b )
    • setBits( a|b ) = setBits(a) + setBits(b) – setBits( a&b )
  • Therefore, the given equation becomes:
    • ( setBits( X|Y ) – setBits( X&Y ) )+( 2 × setBits( X&Y ) ) = M
    • setBits( X ) + setBits ( Y ) – setBits( X&Y ) – setBits( X&Y ) + ( 2 × setBits ( X&Y ) ) = M
    • setBits( X ) + setBits( Y ) = M
  • Therefore, the task reduces to counting the pairs of elements whose sum of set bits is equal to M.

Follow the steps below to solve the problem:

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count number
// of setbits in the number n
int countsetbits(int n)
{
    // Stores the count of setbits
    int count = 0;
 
    // Iterate while N is
    // greater than 0
    while (n) {
 
        // Increment count by 1
        // if N is odd
        count += n & 1;
 
        // Right shift N
        n >>= 1;
    }
 
    // Return the count of set bits
    return (count);
}
 
// Function to find total count of
// given pairs satisfying the equation
int countPairs(int a[], int N, int M)
{
 
    for (int i = 0; i < N; i++) {
 
        // Update arr[i] with the count
        // of set bits of arr[i]
        a[i] = countsetbits(a[i]);
    }
 
    // Stores the frequency
    // of each array element
    unordered_map<int, int> mp;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Increment the count
        // of arr[i] in mp
        mp[a[i]]++;
    }
 
    // Stores the total count of pairs
    int count = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // Increment count by mp[M - a[i]]
        count += mp[M - a[i]];
 
        // If a[i] is equal to M-a[i]
        if (a[i] == M - a[i]) {
 
            // Decrement count by 1
            count--;
        }
    }
 
    // Return count/2
    return (count / 2);
}
 
// Driver Code
int main()
{
    // Input
    int arr[] = { 3, 0, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int M = 2;
 
    cout << countPairs(arr, N, M);
}

                    

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to count number
// of setbits in the number n
static int countsetbits(int n)
{
     
    // Stores the count of setbits
    int count = 0;
 
    // Iterate while N is
    // greater than 0
    while (n != 0)
    {
         
        // Increment count by 1
        // if N is odd
        count += n & 1;
 
        // Right shift N
        n >>= 1;
    }
 
    // Return the count of set bits
    return (count);
}
 
// Function to find total count of
// given pairs satisfying the equation
static int countPairs(int[] a, int N, int M)
{
    for(int i = 0; i < N; i++)
    {
         
        // Update arr[i] with the count
        // of set bits of arr[i]
        a[i] = countsetbits(a[i]);
    }
 
    // Stores the frequency
    // of each array element
     HashMap<Integer,
             Integer> mp = new HashMap<Integer,
                                       Integer>();
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
        if (mp.containsKey(a[i]))
        {
            mp.put(a[i], mp.get(a[i]) + 1);
        }
        else
        {
            mp.put(a[i], 1);
        }
    }
     
    // Stores the total count of pairs
    int count = 0;
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // Increment count by mp[M - a[i]]
        count += mp.get(M - a[i]);
 
        // If a[i] is equal to M-a[i]
        if (a[i] == M - a[i])
        {
             
            // Decrement count by 1
            count--;
        }
    }
 
    // Return count/2
    return (count / 2);
}   
 
// Driver Code
public static void main(String[] args)
{
     
    // Input
    int[] arr = { 3, 0, 4, 5 };
    int N = arr.length;
    int M = 2;
 
    System.out.println(countPairs(arr, N, M));
}
}
 
// This code is contributed by avijitmondal1998

                    

Python3

# Python3 Program for the above approach
 
# Function to count number
# of setbits in the number n
def  countSetBits(n):
   
    # Stores the count of setbits
    count = 0
     
    # Iterate while N is
    # greater than 0
    while (n):
       
        # Increment count by 1
        # if N is odd
        count += n & 1
         
        # Right shift N
        n >>= 1
         
    # Return the count of set bits
    return count
 
def countPairs(arr, N, M):
    for i in range(0, N):
       
        # Update arr[i] with the count
        # of set bits of arr[i]
        arr[i] = countSetBits(arr[i])
         
    # Store counts of all elements in a dictionary
    mp = {}
    for i in range(0, N):
        if arr[i] in mp:
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1
    twice_count = 0
     
    # Iterate through each element and increment
    # the count (Notice that every pair is
    # counted twice)
    for i in range(0, N):
        if M - arr[i] in mp.keys():
            twice_count += mp[M - arr[i]]
             
        # if (arr[i], arr[i]) pair satisfies the
        # condition, then we need to ensure that
        # the count is  decreased by one such
        # that the (arr[i], arr[i]) pair is not
        # considered
        if (M - arr[i] == arr[i]):
            twice_count -= 1
  
    # return the half of twice_count
    return int(twice_count / 2)
         
# Driver code
# Input
arr = [ 3, 0, 4, 5]
N = len(arr)
M = 2
print(countPairs(arr, N, M))
 
# This code is contributed by santhoshcharan.

                    

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count number
// of setbits in the number n
static int countsetbits(int n)
{
     
    // Stores the count of setbits
    int count = 0;
 
    // Iterate while N is
    // greater than 0
    while (n != 0)
    {
         
        // Increment count by 1
        // if N is odd
        count += n & 1;
 
        // Right shift N
        n >>= 1;
    }
 
    // Return the count of set bits
    return (count);
}
 
// Function to find total count of
// given pairs satisfying the equation
static int countPairs(int[] a, int N, int M)
{
    for(int i = 0; i < N; i++)
    {
         
        // Update arr[i] with the count
        // of set bits of arr[i]
        a[i] = countsetbits(a[i]);
    }
 
    // Stores the frequency
    // of each array element
    Dictionary<int,
               int> mp = new Dictionary<int,
                                        int>();
 
    // Traverse the array
    for(int i = 0; i < N; ++i)
    {
            
        // Update frequency of
        // each array element
        if (mp.ContainsKey(a[i]) == true)
            mp[a[i]] += 1;
        else
            mp[a[i]] = 1;
    }
     
    // Stores the total count of pairs
    int count = 0;
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // Increment count by mp[M - a[i]]
        count += mp[M - a[i]];
 
        // If a[i] is equal to M-a[i]
        if (a[i] == M - a[i])
        {
             
            // Decrement count by 1
            count--;
        }
    }
 
    // Return count/2
    return (count / 2);
}
 
// Driver Code
static public void Main()
{
     
    // Input
    int[] arr = { 3, 0, 4, 5 };
    int N = arr.Length;
    int M = 2;
 
    Console.WriteLine(countPairs(arr, N, M));
}
}
 
// This code is contributed by sanjoy_62

                    

Javascript

<script>
// Javascript program for the above approach
 
// Function to count number
// of setbits in the number n
function countsetbits(n)
{
    // Stores the count of setbits
    var count = 0;
 
    // Iterate while N is
    // greater than 0
    while (n) {
 
        // Increment count by 1
        // if N is odd
        count += n & 1;
 
        // Right shift N
        n >>= 1;
    }
 
    // Return the count of set bits
    return (count);
}
 
// Function to find total count of
// given pairs satisfying the equation
function countPairs(a, N, M)
{
 
    for (var i = 0; i < N; i++) {
 
        // Update arr[i] with the count
        // of set bits of arr[i]
        a[i] = countsetbits(a[i]);
    }
 
    // Stores the frequency
    // of each array element
    var mp = new Map();
 
    // Traverse the array
    for (var i = 0; i < N; i++) {
 
        // Increment the count
        // of arr[i] in mp
        if(mp.has(a[i]))
            mp.set(a[i], mp.get(a[i])+1)
        else
            mp.set(a[i], 1)
    }
 
    // Stores the total count of pairs
    var count = 0;
 
    // Traverse the array arr[]
    for (var i = 0; i < N; i++) {
 
        // Increment count by mp[M - a[i]]
        count += mp.get(M - a[i]);
 
        // If a[i] is equal to M-a[i]
        if (a[i] == M - a[i]) {
 
            // Decrement count by 1
            count--;
        }
    }
 
    // Return count/2
    return (count / 2);
}
 
// Driver Code
// Input
var arr = [3, 0, 4, 5];
var N = arr.length;
var M = 2;
document.write( countPairs(arr, N, M));
 
</script>

                    

Output
2





Time Complexity: O(NlogN)
Auxiliary Space: O(N) as using auxiliary space for unordered_map


 



Last Updated : 11 Oct, 2023
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