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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
  • Difficulty Level : Expert
  • Last Updated : 24 Mar, 2021

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.



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]) {
 
            // Decrment 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])
        {
             
            // Decrment 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 cose 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])
        {
             
            // Decrment 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
Output: 
2

 

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

 

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