# Count of pairs in an array such that the highest power of 2 that divides their product is 1

• Last Updated : 28 May, 2022

Given an array arr[] of N positive integers. The task is to find the count of pairs (arr[i], arr[j]) such that the maximum power of 2 that divides arr[i] * arr[j] is 1.
Examples:

Input: arr[] = {3, 5, 2, 8}
Output:
(3, 2), (5, 2) and (3, 5) are the only valid pairs.
Since the power of 2 that divides 3 * 2 = 6 is 1,
5 * 2 = 10 is 1 and 3 * 5 = 15 is 0.
Input: arr[] = {4, 2, 7, 11, 14, 15, 18}
Output: 12

Approach: As the maximum power of 2 that divides arr[i] * arr[j] is at max 1, it means that if P is the product then it must either be odd or 2 is the only even factor of P
It implies that both arr[i] and arr[j] must be odd or exactly one of them is even and 2 is the only even factor of this number.
If odd is the count of odd numbers and even is the count of even numbers such that 2 is the only even factor of that number then the answer will be odd * even + 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 valid pairs``int` `cntPairs(``int` `a[], ``int` `n)``{` `    ``// To store the count of odd numbers and``    ``// the count of even numbers such that 2``    ``// is the only even factor of that number``    ``int` `odd = 0, even = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// If current number is odd``        ``if` `(a[i] % 2 == 1)``            ``odd++;` `        ``// If current number is even and 2``        ``// is the only even factor of it``        ``else` `if` `((a[i] / 2) % 2 == 1)``            ``even++;``    ``}` `    ``// Calculate total number of valid pairs``    ``int` `ans = odd * even + (odd * (odd - 1)) / 2;` `    ``return` `ans;``}` `// Driver code``int` `main()``{` `    ``int` `a[] = { 4, 2, 7, 11, 14, 15, 18 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);` `    ``cout << cntPairs(a, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Function to return the count of valid pairs``static` `int` `cntPairs(``int` `a[], ``int` `n)``{` `    ``// To store the count of odd numbers and``    ``// the count of even numbers such that 2``    ``// is the only even factor of that number``    ``int` `odd = ``0``, even = ``0``;` `    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `        ``// If current number is odd``        ``if` `(a[i] % ``2` `== ``1``)``            ``odd++;` `        ``// If current number is even and 2``        ``// is the only even factor of it``        ``else` `if` `((a[i] / ``2``) % ``2` `== ``1``)``            ``even++;``    ``}` `    ``// Calculate total number of valid pairs``    ``int` `ans = odd * even + (odd * (odd - ``1``)) / ``2``;` `    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `a[] = { ``4``, ``2``, ``7``, ``11``, ``14``, ``15``, ``18` `};``    ``int` `n = a.length;` `    ``System.out.println(cntPairs(a, n));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of valid pairs``def` `cntPairs(a, n) :` `    ``# To store the count of odd numbers and``    ``# the count of even numbers such that 2``    ``# is the only even factor of that number``    ``odd ``=` `0``; even ``=` `0``;` `    ``for` `i ``in` `range``(n) :` `        ``# If current number is odd``        ``if` `(a[i] ``%` `2` `=``=` `1``) :``            ``odd ``+``=` `1``;` `        ``# If current number is even and 2``        ``# is the only even factor of it``        ``elif` `((a[i] ``/` `2``) ``%` `2` `=``=` `1``) :``            ``even ``+``=` `1``;``    ` `    ``# Calculate total number of valid pairs``    ``ans ``=` `odd ``*` `even ``+` `(odd ``*` `(odd ``-` `1``)) ``/``/` `2``;` `    ``return` `ans;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``a ``=` `[ ``4``, ``2``, ``7``, ``11``, ``14``, ``15``, ``18` `];``    ``n ``=` `len``(a);` `    ``print``(cntPairs(a, n));``    ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `// Function to return the count of valid pairs``static` `int` `cntPairs(``int` `[]a, ``int` `n)``{` `    ``// To store the count of odd numbers and``    ``// the count of even numbers such that 2``    ``// is the only even factor of that number``    ``int` `odd = 0, even = 0;` `    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `        ``// If current number is odd``        ``if` `(a[i] % 2 == 1)``            ``odd++;` `        ``// If current number is even and 2``        ``// is the only even factor of it``        ``else` `if` `((a[i] / 2) % 2 == 1)``            ``even++;``    ``}` `    ``// Calculate total number of valid pairs``    ``int` `ans = odd * even + (odd * (odd - 1)) / 2;` `    ``return` `ans;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]a = { 4, 2, 7, 11, 14, 15, 18 };``    ``int` `n = a.Length;` `    ``Console.WriteLine(cntPairs(a, n));``}``}` `// This code is contributed by Ajay KUmar`

## Javascript

 ``

Output:

`12`

Time Complexity: O(n) where n is number of elements in given array. As, we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(1), as we are not using any extra space.

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