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Pairs from an array that satisfy the given condition

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  • Difficulty Level : Basic
  • Last Updated : 12 Sep, 2022
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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:
(2, 4), (2, 5) and (3, 4) are the only valid pairs.

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

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++




// 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);
}

Java




// 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.

Python3




# 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

C#




// 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

PHP




<?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
?>

Javascript




<script>
 
// Javascript implementation of the approach
     
// Function to return the number
// of set bits in n
function setBits(n)
{
    let count = 0;
   
    while (n > 0)
    {
        n = n & (n - 1);
        count++;
    }
    return count;
}
 
// Function to return the count of
// required pairs
function countPairs(a, n)
{
    let count = 0;
   
    for(let i = 0; i < n - 1; i++)
    {
         
        // Set bits for first element
        // of the pair
        let setbits_x = setBits(a[i]);
   
        for(let j = i + 1; j < n; j++)
        {
   
            // Set bits for second element
            // of the pair
            let setbits_y = setBits(a[j]);
   
            // Set bits of the resultant number which is
            // the XOR of both the elements of the pair
            let 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
let a = [ 2, 3, 4, 5, 6 ];
let n = a.length;
 
document.write(countPairs(a, n));
 
// This code is contributed by unknown2108
 
</script>

Output

3

Complexity Analysis:

  • Time Complexity: O(N2logM), where N is the size of the given array and M is the maximum element in the array.
  • Auxiliary Space: O(1), no extra space is required, so it is a constant.

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