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++
// 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:
3
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