# Maximum count of pairwise co-prime and common divisors of two given numbers

• Last Updated : 11 May, 2021

Given an array of pairs arr[] of two numbers {N, M}, the task is to find the maximum count of common divisors for each pair N and M such that every pair between the common divisor are co-prime.

A number x is a common divisor of N and M if, N%x = 0 and M%x = 0
Two numbers are co-prime if their Greatest Common Divisor is 1.

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Examples:

Input: arr[][] = {{12, 18}, {420, 660}}
Output: 3 4
Explanation:
For pair (12, 18):
{1, 2, 3} are common divisors of both 12 and 18, and are pairwise co-prime.
For pair (420, 660):
{1, 2, 3, 5} are common divisors of both 12 and 18, and are pairwise co-prime.
Input: arr[][] = {{8, 18}, {20, 66}}
Output: 2 2

Approach: The maximum count of common divisors of N and M such that the GCD of all the pairs between them is always 1 is 1 and all the common prime divisors of N and M. To count all the common prime divisors the idea is to find the GCD(say G) of the given two numbers and then count the number of prime divisors of the number G.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the gcd of``// two numbers``int` `gcd(``int` `x, ``int` `y)``{``    ``if` `(x % y == 0)``        ``return` `y;``    ``else``        ``return` `gcd(y, x % y);``}` `// Function to of pairwise co-prime``// and common divisors of two numbers``int` `countPairwiseCoprime(``int` `N, ``int` `M)``{``    ``// Initialize answer with 1,``    ``// to include 1 in the count``    ``int` `answer = 1;` `    ``// Count of primes of gcd(N, M)``    ``int` `g = gcd(N, M);``    ``int` `temp = g;` `    ``// Finding prime factors of gcd``    ``for` `(``int` `i = 2; i * i <= g; i++) {` `        ``// Increment count if it is``        ``// divisible by i``        ``if` `(temp % i == 0) {``            ``answer++;` `            ``while` `(temp % i == 0)``                ``temp /= i;``        ``}``    ``}``    ``if` `(temp != 1)``        ``answer++;` `    ``// Return the total count``    ``return` `answer;``}` `void` `countCoprimePair(``int` `arr[][2], ``int` `N)``{` `    ``// Function Call for each pair``    ``// to calculate the count of``    ``// pairwise co-prime divisors``    ``for` `(``int` `i = 0; i < N; i++) {``        ``cout << countPairwiseCoprime(arr[i][0],``                                     ``arr[i][1])``             ``<< ``' '``;``    ``}``}` `// Driver Code``int` `main()``{``    ``// Given array of pairs``    ``int` `arr[][2] = { { 12, 18 }, { 420, 660 } };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// Function Call``    ``countCoprimePair(arr, N);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG{` `// Function to find the gcd of``// two numbers``static` `int` `gcd(``int` `x, ``int` `y)``{``    ``if` `(x % y == ``0``)``        ``return` `y;``    ``else``        ``return` `gcd(y, x % y);``}` `// Function to of pairwise co-prime``// and common divisors of two numbers``static` `int` `countPairwiseCoprime(``int` `N, ``int` `M)``{``    ``// Initialize answer with 1,``    ``// to include 1 in the count``    ``int` `answer = ``1``;` `    ``// Count of primes of gcd(N, M)``    ``int` `g = gcd(N, M);``    ``int` `temp = g;` `    ``// Finding prime factors of gcd``    ``for` `(``int` `i = ``2``; i * i <= g; i++)``    ``{` `        ``// Increment count if it is``        ``// divisible by i``        ``if` `(temp % i == ``0``)``        ``{``            ``answer++;` `            ``while` `(temp % i == ``0``)``                ``temp /= i;``        ``}``    ``}``    ``if` `(temp != ``1``)``        ``answer++;` `    ``// Return the total count``    ``return` `answer;``}` `static` `void` `countCoprimePair(``int` `arr[][], ``int` `N)``{` `    ``// Function Call for each pair``    ``// to calculate the count of``    ``// pairwise co-prime divisors``    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``        ``System.out.print(countPairwiseCoprime(arr[i][``0``],``                                               ``arr[i][``1``]) + ``" "``);``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``// Given array of pairs``    ``int` `arr[][] = { { ``12``, ``18` `}, { ``420``, ``660` `} };``    ``int` `N = arr.length;` `    ``// Function Call``    ``countCoprimePair(arr, N);``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program for the above approach` `# Function to find the gcd of``# two numbers``def` `gcd(x, y):``    ``if` `(x ``%` `y ``=``=` `0``):``        ``return` `y``    ``else``:``        ``return` `gcd(y, x ``%` `y)` `# Function to of pairwise co-prime``# and common divisors of two numbers``def` `countPairwiseCoprime(N, M):` `    ``# Initialize answer with 1,``    ``# to include 1 in the count``    ``answer ``=` `1` `    ``# Count of primes of gcd(N, M)``    ``g ``=` `gcd(N, M)``    ``temp ``=` `g` `    ``# Finding prime factors of gcd``    ``for` `i ``in` `range``(``2``, g ``+` `1``):` `        ``if` `i ``*` `i > g:``            ``break` `        ``# Increment count if it is``        ``# divisible by i``        ``if` `(temp ``%` `i ``=``=` `0``) :``            ``answer ``+``=` `1` `            ``while` `(temp ``%` `i ``=``=` `0``):``                ``temp ``/``/``=` `i` `    ``if` `(temp !``=` `1``):``        ``answer ``+``=` `1` `    ``# Return the total count``    ``return` `answer` `def` `countCoprimePair(arr, N):` `    ``# Function Call for each pair``    ``# to calculate the count of``    ``# pairwise co-prime divisors``    ``for` `i ``in` `range``(N):``        ``print``(countPairwiseCoprime(arr[i][``0``],``                                   ``arr[i][``1``]),``                                     ``end ``=` `" "``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``  ` `    ``# Given array of pairs``    ``arr``=` `[ [ ``12``, ``18` `], [ ``420``, ``660` `] ]``    ``N ``=` `len``(arr)` `    ``# Function Call``    ``countCoprimePair(arr, N)` `# This code is contributed by Mohit Kumar`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG{` `// Function to find the gcd of``// two numbers``static` `int` `gcd(``int` `x, ``int` `y)``{``    ``if` `(x % y == 0)``        ``return` `y;``    ``else``        ``return` `gcd(y, x % y);``}` `// Function to of pairwise co-prime``// and common divisors of two numbers``static` `int` `countPairwiseCoprime(``int` `N, ``int` `M)``{``    ``// Initialize answer with 1,``    ``// to include 1 in the count``    ``int` `answer = 1;` `    ``// Count of primes of gcd(N, M)``    ``int` `g = gcd(N, M);``    ``int` `temp = g;` `    ``// Finding prime factors of gcd``    ``for` `(``int` `i = 2; i * i <= g; i++)``    ``{` `        ``// Increment count if it is``        ``// divisible by i``        ``if` `(temp % i == 0)``        ``{``            ``answer++;` `            ``while` `(temp % i == 0)``                ``temp /= i;``        ``}``    ``}``    ``if` `(temp != 1)``        ``answer++;` `    ``// Return the total count``    ``return` `answer;``}` `static` `void` `countCoprimePair(``int` `[,]arr, ``int` `N)``{` `    ``// Function Call for each pair``    ``// to calculate the count of``    ``// pairwise co-prime divisors``    ``for` `(``int` `i = 0; i < N; i++)``    ``{``        ``Console.Write(countPairwiseCoprime(arr[i, 0],``                                           ``arr[i, 1]) + ``" "``);``    ``}``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``// Given array of pairs``    ``int` `[,]arr = { { 12, 18 }, { 420, 660 } };``    ``int` `N = arr.GetLength(0);` `    ``// Function Call``    ``countCoprimePair(arr, N);``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:
`3 4`

Time Complexity: O(X*(sqrt(N) + sqrt(M))), where X is the number of pairs and N & M are two pairs in arr[].
Auxiliary Space: O(1)

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