Pairs from an array that satisfy the given condition

Given an array arr[], the task is to count all the valid pairs from the array. A pair (arr[i], arr[j]) is said to be valid if func( arr[i] ) + func( arr[j] ) = func( XOR(arr[i], arr[j]) ) where func(x) returns the number of set bits in x.

Examples:

Input: arr[] = {2, 3, 4, 5, 6}
Output: 3
(2, 4), (2, 5) and (3, 4) are the only valid pairs.



Input: arr[] = {12, 13, 34, 25, 6}
Output: 4

Approach: Iterating every possible pair and check whether the pair satisfies the given condition. If the condition is satisfied then update count = count + 1. Print the count in the end.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the number
// of set bits in n
int setBits(int n)
{
    int count = 0;
  
    while (n) {
        n = n & (n - 1);
        count++;
    }
    return count;
}
  
// Function to return the count of required pairs
int countPairs(int a[], int n)
{
    int count = 0;
  
    for (int i = 0; i < n - 1; i++) {
  
        // Set bits for first element of the pair
        int setbits_x = setBits(a[i]);
  
        for (int j = i + 1; j < n; j++) {
  
            // Set bits for second element of the pair
            int setbits_y = setBits(a[j]);
  
            // Set bits of the resultant number which is
            // the XOR of both the elements of the pair
            int setbits_xor_xy = setBits(a[i] ^ a[j]);
  
            // If the condition is satisfied
            if (setbits_x + setbits_y == setbits_xor_xy)
  
                // Increment the count
                count++;
        }
    }
  
    // Return the total count
    return count;
}
  
// Driver code
int main()
{
    int a[] = { 2, 3, 4, 5, 6 };
  
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << countPairs(a, n);
}

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Java

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// Java implementation of the approach
import java.io.*;
  
class GFG 
{
      
// Function to return the number
// of set bits in n
static int setBits(int n)
{
    int count = 0;
  
    while (n > 0
    {
        n = n & (n - 1);
        count++;
    }
    return count;
}
  
// Function to return the count of 
// required pairs
static int countPairs(int a[], int n)
{
    int count = 0;
  
    for (int i = 0; i < n - 1; i++)
    {
  
        // Set bits for first element 
        // of the pair
        int setbits_x = setBits(a[i]);
  
        for (int j = i + 1; j < n; j++) 
        {
  
            // Set bits for second element 
            // of the pair
            int setbits_y = setBits(a[j]);
  
            // Set bits of the resultant number which is
            // the XOR of both the elements of the pair
            int setbits_xor_xy = setBits(a[i] ^ a[j]);
  
            // If the condition is satisfied
            if (setbits_x + setbits_y == setbits_xor_xy)
  
                // Increment the count
                count++;
        }
    }
  
    // Return the total count
    return count;
}
  
    // Driver code
    public static void main (String[] args)
    {
  
        int []a = { 2, 3, 4, 5, 6 };
        int n = a.length;
        System.out.println(countPairs(a, n));
    }
}
  
// This code is contributed by ajit.

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Python3

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# Python 3 implementation of the approach
  
# Function to return the number
# of set bits in n
def setBits(n):
    count = 0
  
    while (n):
        n = n & (n - 1)
        count += 1
  
    return count
  
# Function to return the count
# of required pairs
def countPairs(a, n):
    count = 0
  
    for i in range(0, n - 1, 1):
          
        # Set bits for first element 
        # of the pair
        setbits_x = setBits(a[i])
  
        for j in range(i + 1, n, 1):
              
            # Set bits for second element 
            # of the pair
            setbits_y = setBits(a[j])
  
            # Set bits of the resultant number
            # which is the XOR of both the 
            # elements of the pair
            setbits_xor_xy = setBits(a[i] ^ a[j]);
  
            # If the condition is satisfied
            if (setbits_x + 
                setbits_y == setbits_xor_xy):
                  
                # Increment the count
                count += 1
  
    # Return the total count
    return count
  
# Driver code
if __name__ == '__main__':
    a = [2, 3, 4, 5, 6]
  
    n = len(a)
    print(countPairs(a, n))
  
# This code is contributed by
# Sanjit_Prasad

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C#

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// C# implementation of the approach
using System;
class GFG
{
  
// Function to return the number
// of set bits in n
static int setBits(int n)
{
    int count = 0;
  
    while (n > 0) 
    {
        n = n & (n - 1);
        count++;
    }
    return count;
}
  
// Function to return the count of 
// required pairs
static int countPairs(int []a, int n)
{
    int count = 0;
  
    for (int i = 0; i < n - 1; i++)
    {
  
        // Set bits for first element 
        // of the pair
        int setbits_x = setBits(a[i]);
  
        for (int j = i + 1; j < n; j++) 
        {
  
            // Set bits for second element 
            // of the pair
            int setbits_y = setBits(a[j]);
  
            // Set bits of the resultant number which is
            // the XOR of both the elements of the pair
            int setbits_xor_xy = setBits(a[i] ^ a[j]);
  
            // If the condition is satisfied
            if (setbits_x + setbits_y == setbits_xor_xy)
  
                // Increment the count
                count++;
        }
    }
  
    // Return the total count
    return count;
}
  
// Driver code
public static void Main()
{
    int []a = { 2, 3, 4, 5, 6 };
  
    int n = a.Length;
  
    Console.Write(countPairs(a, n));
}
}
  
// This code is contributed 
// by Akanksha Rai

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PHP

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<?php 
// PHP implementation of the approach
  
// Function to return the number
// of set bits in n
function setBits($n)
{
    $count = 0;
  
    while ($n
    {
        $n = $n & ($n - 1);
        $count++;
    }
    return $count;
}
  
// Function to return the count of 
// required pairs
function countPairs(&$a, $n)
{
    $count = 0;
  
    for ($i = 0; $i < $n - 1; $i++) 
    {
  
        // Set bits for first element 
        // of the pair
        $setbits_x = setBits($a[$i]);
  
        for ($j = $i + 1; $j < $n; $j++) 
        {
  
            // Set bits for second element of the pair
            $setbits_y = setBits($a[$j]);
  
            // Set bits of the resultant number which is
            // the XOR of both the elements of the pair
            $setbits_xor_xy = setBits($a[$i] ^ $a[$j]);
  
            // If the condition is satisfied
            if ($setbits_x
                $setbits_y == $setbits_xor_xy)
  
                // Increment the count
                $count++;
        }
    }
  
    // Return the total count
    return $count;
}
  
// Driver code
$a = array(2, 3, 4, 5, 6 );
$n = sizeof($a) / sizeof($a[0]);
echo countPairs($a, $n);
  
// This code is contributed by ita_c
?>

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Output:

3


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