# 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

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

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 ` `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); ` `    ``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

 ` `

Output:

```3
```

Time Complexity: O(N)

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