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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 
?> 

Output:

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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