Composite XOR and Coprime AND

Given an array arr[], the task is to count the number of unordered pairs of indices (i, j) such the gcd(2, a[i]^a[j]) > 1 and gcd(2, a[i] & a[j]) < 2 where ^ and & are bitwise XOR and bitwise AND operations respectively.
Examples:

Input: arr[] = {1, 2, 3, 4, 5} 
Output: 3 
All valid pairs are (1, 3), (1, 5) and (3, 5)
Input: arr[] = {21, 121, 13, 44, 65} 
Output: 6 

Approach:

• gcd(2, a[i]^a[j]) > 1 will only be true if both a[i] and a[j] are either even or odd.
• Narrowing down the pairs from the previous step, a pair (a, b) will never yield gcd(2, a[i] & a[j]) < 2 if both a and b are even. So, only pairs where a and b are both odd will satisfy both the given conditions.
• Count the number of odd element sin the given array and store it ion odd and the result will be (odd * (odd – 1)) / 2.

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 count of
// odd numbers in the array
int countOdd(int arr[], int n)
{

    // Variable to count odd numbers
    int odd = 0;

    for (int i = 0; i < n; i++) {

        // Odd number
        if (arr[i] % 2 == 1)
            odd++;
    }

    return odd;
}

// Function to return the count of valid pairs
int countValidPairs(int arr[], int n)
{
    int odd = countOdd(arr, n);

    return (odd * (odd - 1)) / 2;
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countValidPairs(arr, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    // Function to return the count of
    // odd numbers in the array
    static int countOdd(int [] arr, int n)
    {

        // Variable to count odd numbers
        int odd = 0;

        for (int i = 0; i < n; i++)
        {

            // Odd number
            if (arr[i] % 2 == 1)
                odd++;
        }
        return odd;
    }

    // Function to return the count of valid pairs
    static int countValidPairs(int [] arr, int n)
    {
        int odd = countOdd(arr, n);

        return (odd * (odd - 1)) / 2;
    }

    // Driver Code
    public static void main(String []args)
    {
        int [] arr = { 1, 2, 3, 4, 5 };
        int n = arr.length;
        System.out.println(countValidPairs(arr, n));
    }
}

// This code is contributed by ihritik

Python3

# Python3 implementation of the approach

# Function to return the count of
# odd numbers in the array
def countOdd(arr, n):

    # Variable to count odd numbers
    odd = 0;

    for i in range(0, n):

        # Odd number
        if (arr[i] % 2 == 1):
            odd = odd + 1;

    return odd;

# Function to return the count
# of valid pairs
def countValidPairs(arr, n):

    odd = countOdd(arr, n);

    return (odd * (odd - 1)) / 2;

# Driver Code
arr = [1, 2, 3, 4, 5 ];
n = len(arr);
print(int(countValidPairs(arr, n)));

# This code is contributed by
# Shivi_Aggarwal

C#

// C# implementation of the approach
using System;

class GFG
{
    // Function to return the count of
    // odd numbers in the array
    static int countOdd(int [] arr, int n)
    {

        // Variable to count odd numbers
        int odd = 0;

        for (int i = 0; i < n; i++)
        {

            // Odd number
            if (arr[i] % 2 == 1)
                odd++;
        }
        return odd;
    }

    // Function to return the count of valid pairs
    static int countValidPairs(int [] arr, int n)
    {
        int odd = countOdd(arr, n);

        return (odd * (odd - 1)) / 2;
    }

    // Driver Code
    public static void Main()
    {
        int [] arr = { 1, 2, 3, 4, 5 };
        int n = arr.Length;
        Console.WriteLine(countValidPairs(arr, n));
    }
}

// This code is contributed by ihritik

PHP

<?php
// PHP implementation of the approach

// Function to return the count of
// odd numbers in the array
function countOdd($arr, $n)
{

    // Variable to count odd numbers
    $odd = 0;

    for ($i = 0; $i < $n; $i++)
    {

        // Odd number
        if ($arr[$i] % 2 == 1)
            $odd++;
    }

    return $odd;
}

// Function to return the count
// of valid pairs
function countValidPairs($arr, $n)
{
    $odd = countOdd($arr, $n);

    return ($odd * ($odd - 1)) / 2;
}

// Driver Code
$arr = array(1, 2, 3, 4, 5);
$n = sizeof($arr);
echo countValidPairs($arr, $n);

// This code is contributed by Ryuga
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to return the count of
// odd numbers in the array
function countOdd(arr, n)
{

    // Variable to count odd numbers
    var odd = 0;

    for(var i = 0; i < n; i++)
    {

        // Odd number
        if (arr[i] % 2 == 1)
            odd++;
    }
    return odd;
}

// Function to return the count of valid pairs
function countValidPairs(arr, n)
{
    var odd = countOdd(arr, n);

    return (odd * (odd - 1)) / 2;
}

// Driver Code
var arr = [ 1, 2, 3, 4, 5 ];
var n = arr.length;

document.write(countValidPairs(arr, n));

// This code is contributed by Khushboogoyal499

</script>

Output:

3

Time Complexity: O(N)

Auxiliary Space: O(1)