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# Common Divisors of Two Numbers

Given two integer numbers, the task is to find count of all common divisors of given numbers?

Examples :

```Input : a = 12, b = 24
Output: 6
// all common divisors are 1, 2, 3,
// 4, 6 and 12

Input : a = 3, b = 17
Output: 1
// all common divisors are 1

Input : a = 20, b = 36
Output: 3
// all common divisors are 1, 2, 4```
Recommended Practice

It is recommended to refer all divisors of a given number as a prerequisite of this article.

Naive Solution
A simple solution is to first find all divisors of first number and store them in an array or hash. Then find common divisors of second number and store them. Finally print common elements of two stored arrays or hash. The key is that the magnitude of powers of prime factors of a divisor should be equal to the minimum power of two prime factors of a and b.

• Find the prime factors of a using prime factorization.
• Find the count of each prime factor of a and store it in a Hashmap.
• Prime factorize b using distinct prime factors of a.
• Then the total number of divisors would be equal to the product of (count + 1)
of each factor.
• count is the minimum of counts of each prime factors of a and b.
• This gives the count of all divisors of a and b.

## C++

 `// C++ implementation of program``#include ``using` `namespace` `std;` `// Map to store the count of each``// prime factor of a``map<``int``, ``int``> ma;` `// Function that calculate the count of``// each prime factor of a number``void` `primeFactorize(``int` `a)``{``    ``for``(``int` `i = 2; i * i <= a; i += 2)``    ``{``        ``int` `cnt = 0;``        ``while` `(a % i == 0)``        ``{``            ``cnt++;``            ``a /= i;``        ``}``        ``ma[i] = cnt;``    ``}``    ``if` `(a > 1)``    ``{``        ``ma[a] = 1;``    ``}``}` `// Function to calculate all common``// divisors of two given numbers``// a, b --> input integer numbers``int` `commDiv(``int` `a, ``int` `b)``{``    ` `    ``// Find count of each prime factor of a``    ``primeFactorize(a);` `    ``// stores number of common divisors``    ``int` `res = 1;` `    ``// Find the count of prime factors``    ``// of b using distinct prime factors of a``    ``for``(``auto` `m = ma.begin();``             ``m != ma.end(); m++)``    ``{``        ``int` `cnt = 0;``        ``int` `key = m->first;``        ``int` `value = m->second;` `        ``while` `(b % key == 0)``        ``{``            ``b /= key;``            ``cnt++;``        ``}` `        ``// Prime factor of common divisor``        ``// has minimum cnt of both a and b``        ``res *= (min(cnt, value) + 1);``    ``}``    ``return` `res;``}` `// Driver code   ``int` `main()``{``    ``int` `a = 12, b = 24;``    ` `    ``cout << commDiv(a, b) << endl;``    ` `    ``return` `0;``}` `// This code is contributed by divyeshrabadiya07`

## Java

 `// Java implementation of program``import` `java.util.*;``import` `java.io.*;` `class` `GFG {``    ``// map to store the count of each prime factor of a``    ``static` `HashMap ma = ``new` `HashMap<>();` `    ``// method that calculate the count of``    ``// each prime factor of a number``    ``static` `void` `primeFactorize(``int` `a)``    ``{``        ``for` `(``int` `i = ``2``; i * i <= a; i += ``2``) {``            ``int` `cnt = ``0``;``            ``while` `(a % i == ``0``) {``                ``cnt++;``                ``a /= i;``            ``}``            ``ma.put(i, cnt);``        ``}``        ``if` `(a > ``1``)``            ``ma.put(a, ``1``);``    ``}` `    ``// method to calculate all common divisors``    ``// of two given numbers``    ``// a, b --> input integer numbers``    ``static` `int` `commDiv(``int` `a, ``int` `b)``    ``{``        ``// Find count of each prime factor of a``        ``primeFactorize(a);` `        ``// stores number of common divisors``        ``int` `res = ``1``;` `        ``// Find the count of prime factors of b using``        ``// distinct prime factors of a``        ``for` `(Map.Entry m : ma.entrySet()) {``            ``int` `cnt = ``0``;` `            ``int` `key = m.getKey();``            ``int` `value = m.getValue();` `            ``while` `(b % key == ``0``) {``                ``b /= key;``                ``cnt++;``            ``}` `            ``// prime factor of common divisor``            ``// has minimum cnt of both a and b``            ``res *= (Math.min(cnt, value) + ``1``);``        ``}``        ``return` `res;``    ``}` `    ``// Driver method``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `a = ``12``, b = ``24``;``        ``System.out.println(commDiv(a, b));``    ``}``}`

## Python3

 `# Python3 implementation of program``import` `math` `# Map to store the count of each``# prime factor of a``ma ``=` `{}` `# Function that calculate the count of``# each prime factor of a number``def` `primeFactorize(a):``    ` `    ``sqt ``=` `int``(math.sqrt(a))``    ``for` `i ``in` `range``(``2``, sqt, ``2``):``        ``cnt ``=` `0``        ` `        ``while` `(a ``%` `i ``=``=` `0``):``            ``cnt ``+``=` `1``            ``a ``/``=` `i``            ` `        ``ma[i] ``=` `cnt``        ` `    ``if` `(a > ``1``):``        ``ma[a] ``=` `1``        ` `# Function to calculate all common``# divisors of two given numbers``# a, b --> input integer numbers``def` `commDiv(a, b):``    ` `    ``# Find count of each prime factor of a``    ``primeFactorize(a)``    ` `    ``# stores number of common divisors``    ``res ``=` `1``    ` `    ``# Find the count of prime factors``    ``# of b using distinct prime factors of a``    ``for` `key, value ``in` `ma.items():``        ``cnt ``=` `0``        ` `        ``while` `(b ``%` `key ``=``=` `0``):``            ``b ``/``=` `key``            ``cnt ``+``=` `1``            ` `        ``# Prime factor of common divisor``        ``# has minimum cnt of both a and b``        ``res ``*``=` `(``min``(cnt, value) ``+` `1``)``        ` `    ``return` `res``    ` `# Driver code   ``a ``=` `12``b ``=` `24` `print``(commDiv(a, b))` `# This code is contributed by Stream_Cipher`

## C#

 `// C# implementation of program``using` `System;``using` `System.Collections.Generic;` `class` `GFG{``  ` `// Map to store the count of each``// prime factor of a``static` `Dictionary<``int``,``                  ``int``> ma = ``new` `Dictionary<``int``,``                                           ``int``>();` `// Function that calculate the count of``// each prime factor of a number``static` `void` `primeFactorize(``int` `a)``{``    ``for``(``int` `i = 2; i * i <= a; i += 2)``    ``{``        ``int` `cnt = 0;``        ``while` `(a % i == 0)``        ``{``            ``cnt++;``            ``a /= i;``        ``}``        ``ma.Add(i, cnt);``    ``}``    ` `    ``if` `(a > 1)``        ``ma.Add(a, 1);``}` `// Function to calculate all common``// divisors of two given numbers``// a, b --> input integer numbers``static` `int` `commDiv(``int` `a, ``int` `b)``{``    ` `    ``// Find count of each prime factor of a``    ``primeFactorize(a);``    ` `    ``// Stores number of common divisors``    ``int` `res = 1;``    ` `    ``// Find the count of prime factors``    ``// of b using distinct prime factors of a``    ``foreach``(KeyValuePair<``int``, ``int``> m ``in` `ma)``    ``{``        ``int` `cnt = 0;``        ``int` `key = m.Key;``        ``int` `value = m.Value;``        ` `        ``while` `(b % key == 0)``        ``{``            ``b /= key;``            ``cnt++;``        ``}` `        ``// Prime factor of common divisor``        ``// has minimum cnt of both a and b``        ``res *= (Math.Min(cnt, value) + 1);``    ``}``    ``return` `res;``}` `// Driver code   ``static` `void` `Main()``{``    ``int` `a = 12, b = 24;``    ` `    ``Console.WriteLine(commDiv(a, b));``}``}` `// This code is contributed by divyesh072019`

## Javascript

 ``

Output:

`6`

Time Complexity: O(√n log n)
Auxiliary Space: O(n)

Efficient Solution –
A better solution is to calculate the greatest common divisor (gcd) of given two numbers, and then count divisors of that gcd.

## C++

 `// C++ implementation of program``#include ``using` `namespace` `std;` `// Function to calculate gcd of two numbers``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to calculate all common divisors``// of two given numbers``// a, b --> input integer numbers``int` `commDiv(``int` `a, ``int` `b)``{``    ``// find gcd of a, b``    ``int` `n = gcd(a, b);` `    ``// Count divisors of n.``    ``int` `result = 0;``    ``for` `(``int` `i = 1; i <= ``sqrt``(n); i++) {``        ``// if 'i' is factor of n``        ``if` `(n % i == 0) {``            ``// check if divisors are equal``            ``if` `(n / i == i)``                ``result += 1;``            ``else``                ``result += 2;``        ``}``    ``}``    ``return` `result;``}` `// Driver program to run the case``int` `main()``{``    ``int` `a = 12, b = 24;``    ``cout << commDiv(a, b);``    ``return` `0;``}`

## Java

 `// Java implementation of program` `class` `Test {``    ``// method to calculate gcd of two numbers``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == ``0``)``            ``return` `b;` `        ``return` `gcd(b % a, a);``    ``}``    ``// method to calculate all common divisors``    ``// of two given numbers``    ``// a, b --> input integer numbers``    ``static` `int` `commDiv(``int` `a, ``int` `b)``    ``{``        ``// find gcd of a, b``        ``int` `n = gcd(a, b);` `        ``// Count divisors of n.``        ``int` `result = ``0``;``        ``for` `(``int` `i = ``1``; i <= Math.sqrt(n); i++) {``            ``// if 'i' is factor of n``            ``if` `(n % i == ``0``) {``                ``// check if divisors are equal``                ``if` `(n / i == i)``                    ``result += ``1``;``                ``else``                    ``result += ``2``;``            ``}``        ``}``        ``return` `result;``    ``}` `    ``// Driver method``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `a = ``12``, b = ``24``;``        ``System.out.println(commDiv(a, b));``    ``}``}`

## Python3

 `# Python implementation of program``from` `math ``import` `sqrt`  `# Function to calculate gcd of two numbers``def` `gcd(a, b):``    ` `    ``if` `a ``=``=` `0``:``        ``return` `b``    ``return` `gcd(b ``%` `a, a)``  ` `# Function to calculate all common divisors``# of two given numbers``# a, b --> input integer numbers``def` `commDiv(a, b):``    ` `    ``# find GCD of a, b``    ``n ``=` `gcd(a, b)` `    ``# Count divisors of n``    ``result ``=` `0``    ``for` `i ``in` `range``(``1``,``int``(sqrt(n))``+``1``):` `        ``# if i is a factor of n``        ``if` `n ``%` `i ``=``=` `0``:` `            ``# check if divisors are equal``            ``if` `n``/``i ``=``=` `i:``                ``result ``+``=` `1``            ``else``:``                ``result ``+``=` `2``                ` `    ``return` `result` `# Driver program to run the case``if` `__name__ ``=``=` `"__main__"``:``    ``a ``=` `12``    ``b ``=` `24``;``    ``print``(commDiv(a, b))`

## C#

 `// C# implementation of program``using` `System;` `class` `GFG {` `    ``// method to calculate gcd``    ``// of two numbers``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == 0)``            ``return` `b;` `        ``return` `gcd(b % a, a);``    ``}` `    ``// method to calculate all``    ``// common divisors of two``    ``// given numbers a, b -->``    ``// input integer numbers``    ``static` `int` `commDiv(``int` `a, ``int` `b)``    ``{` `        ``// find gcd of a, b``        ``int` `n = gcd(a, b);` `        ``// Count divisors of n.``        ``int` `result = 0;``        ``for` `(``int` `i = 1; i <= Math.Sqrt(n); i++) {` `            ``// if 'i' is factor of n``            ``if` `(n % i == 0) {` `                ``// check if divisors are equal``                ``if` `(n / i == i)``                    ``result += 1;``                ``else``                    ``result += 2;``            ``}``        ``}` `        ``return` `result;``    ``}` `    ``// Driver method``    ``public` `static` `void` `Main(String[] args)``    ``{` `        ``int` `a = 12, b = 24;` `        ``Console.Write(commDiv(a, b));``    ``}``}` `// This code contributed by parashar.`

## PHP

 ` input integer numbers``function` `commDiv(``\$a``, ``\$b``)``{``    ``// find gcd of a, b``    ``\$n` `= gcd(``\$a``, ``\$b``);` `    ``// Count divisors of n.``    ``\$result` `= 0;``    ``for` `(``\$i` `= 1; ``\$i` `<= sqrt(``\$n``);``                 ``\$i``++)``    ``{``        ``// if 'i' is factor of n``        ``if` `(``\$n` `% ``\$i` `== 0)``        ``{``            ``// check if divisors``            ``// are equal``            ``if` `(``\$n` `/ ``\$i` `== ``\$i``)``                ``\$result` `+= 1;``            ``else``                ``\$result` `+= 2;``        ``}``    ``}``    ``return` `\$result``;``}` `// Driver Code``\$a` `= 12; ``\$b` `= 24;``echo``(commDiv(``\$a``, ``\$b``));` `// This code is contributed by Ajit.``?>`

## Javascript

 ``

Output :

`6`

Time complexity: O(n1/2) where n is the gcd of two numbers.
Auxiliary Space: O(1)

This article is contributed by Aarti_Rathi and Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

#### Another Approach:

1. Define a function “gcd” that takes two integers “a” and “b” and returns their greatest common divisor (GCD) using the Euclidean algorithm.
2. Define a function “count_common_divisors” that takes two integers “a” and “b” and counts the number of common divisors of “a” and “b” using their GCD.
3. Calculate the GCD of “a” and “b” using the “gcd” function.
4. Initialize a counter “count” to 0.
5. Loop through all possible divisors of the GCD of “a” and “b” from 1 to the square root of the GCD.
6. If the current divisor divides the GCD evenly, increment the counter by 2 (because both “a” and “b” are divisible by the divisor).
7. If the square of the current divisor equals the GCD, decrement the counter by 1 (because we’ve already counted this divisor once).
8. Return the final count of common divisors.
9. In the main function, define two integers “a” and “b” and call the “count_common_divisors” function with these integers.
10. Print the number of common divisors of “a” and “b” using the printf function.

## C

 `#include ` `int` `gcd(``int` `a, ``int` `b) {``    ``if``(b == 0) {``        ``return` `a;``    ``}``    ``return` `gcd(b, a % b);``}` `int` `count_common_divisors(``int` `a, ``int` `b) {``    ``int` `gcd_ab = gcd(a, b);``    ``int` `count = 0;` `    ``for``(``int` `i = 1; i * i <= gcd_ab; i++) {``        ``if``(gcd_ab % i == 0) {``            ``count += 2;``            ``if``(i * i == gcd_ab) {``                ``count--;``            ``}``        ``}``    ``}` `    ``return` `count;``}` `int` `main() {``    ``int` `a = 12;``    ``int` `b = 18;` `    ``int` `common_divisors = count_common_divisors(a, b);` `    ``printf``(``"The number of common divisors of %d and %d is %d.\n"``, a, b, common_divisors);` `    ``return` `0;``}`

## C++

 `#include ``using` `namespace` `std;` `int` `gcd(``int` `a, ``int` `b) {``    ``if``(b == 0) {``        ``return` `a;``    ``}``    ``return` `gcd(b, a % b);``}` `int` `count_common_divisors(``int` `a, ``int` `b) {``    ``int` `gcd_ab = gcd(a, b);``    ``int` `count = 0;` `    ``for``(``int` `i = 1; i * i <= gcd_ab; i++) {``        ``if``(gcd_ab % i == 0) {``            ``count += 2;``            ``if``(i * i == gcd_ab) {``                ``count--;``            ``}``        ``}``    ``}` `    ``return` `count;``}` `int` `main() {``    ``int` `a = 12;``    ``int` `b = 18;` `    ``int` `common_divisors = count_common_divisors(a, b);``    ``cout<<``"The number of common divisors of "``<

## Java

 `import` `java.util.*;` `public` `class` `Main {` `  ``public` `static` `int` `gcd(``int` `a, ``int` `b) {``    ``if``(b == ``0``) {``      ``return` `a;``    ``}``    ``return` `gcd(b, a % b);``  ``}` `  ``public` `static` `int` `countCommonDivisors(``int` `a, ``int` `b) {``    ``int` `gcd_ab = gcd(a, b);``    ``int` `count = ``0``;` `    ``for``(``int` `i = ``1``; i * i <= gcd_ab; i++) {``      ``if``(gcd_ab % i == ``0``) {``        ``count += ``2``;``        ``if``(i * i == gcd_ab) {``          ``count--;``        ``}``      ``}``    ``}` `    ``return` `count;``  ``}` `  ``public` `static` `void` `main(String[] args) {``    ``int` `a = ``12``;``    ``int` `b = ``18``;` `    ``int` `commonDivisors = countCommonDivisors(a, b);``    ``System.out.println(``"The number of common divisors of "` `+ a + ``" and "` `+ b + ``" is "` `+ commonDivisors + ``"."``);``  ``}``}`

## Python3

 `import` `math` `def` `gcd(a, b):``    ``if` `b ``=``=` `0``:``        ``return` `a``    ``return` `gcd(b, a ``%` `b)` `def` `count_common_divisors(a, b):``    ``gcd_ab ``=` `gcd(a, b)``    ``count ``=` `0` `    ``for` `i ``in` `range``(``1``, ``int``(math.sqrt(gcd_ab)) ``+` `1``):``        ``if` `gcd_ab ``%` `i ``=``=` `0``:``            ``count ``+``=` `2``            ``if` `i ``*` `i ``=``=` `gcd_ab:``                ``count ``-``=` `1` `    ``return` `count` `a ``=` `12``b ``=` `18` `common_divisors ``=` `count_common_divisors(a, b)``print``(``"The number of common divisors of"``, a, ``"and"``, b, ``"is"``, common_divisors, ``"."``)` `# This code is contributed by Prajwal Kandekar`

## C#

 `using` `System;` `public` `class` `MainClass``{``    ``public` `static` `int` `GCD(``int` `a, ``int` `b)``    ``{``        ``if` `(b == 0)``        ``{``            ``return` `a;``        ``}``        ``return` `GCD(b, a % b);``    ``}` `    ``public` `static` `int` `CountCommonDivisors(``int` `a, ``int` `b)``    ``{``        ``int` `gcd_ab = GCD(a, b);``        ``int` `count = 0;` `        ``for` `(``int` `i = 1; i * i <= gcd_ab; i++)``        ``{``            ``if` `(gcd_ab % i == 0)``            ``{``                ``count += 2;``                ``if` `(i * i == gcd_ab)``                ``{``                    ``count--;``                ``}``            ``}``        ``}` `        ``return` `count;``    ``}` `    ``public` `static` `void` `Main()``    ``{``        ``int` `a = 12;``        ``int` `b = 18;` `        ``int` `commonDivisors = CountCommonDivisors(a, b);``        ``Console.WriteLine(``"The number of common divisors of {0} and {1} is {2}."``, a, b, commonDivisors);``    ``}``}`

## Javascript

 `// Function to calculate the greatest common divisor of``// two integers a and b using the Euclidean algorithm``function` `gcd(a, b) {``    ``if``(b === 0) {``        ``return` `a;``    ``}``    ``return` `gcd(b, a % b);``}` `// Function to count the number of common divisors of two integers a and b``function` `count_common_divisors(a, b) {``    ``let gcd_ab = gcd(a, b);``    ``let count = 0;` `    ``for``(let i = 1; i * i <= gcd_ab; i++) {``        ``if``(gcd_ab % i === 0) {``            ``count += 2;``            ``if``(i * i === gcd_ab) {``                ``count--;``            ``}``        ``}``    ``}` `    ``return` `count;``}` `let a = 12;``let b = 18;` `let common_divisors = count_common_divisors(a, b);``console.log(`The number of common divisors of \${a} and \${b} is \${common_divisors}.`);`

Output

`The number of common divisors of 12 and 18 is 4.`

The time complexity of the gcd() function is O(log(min(a, b))), as it uses Euclid’s algorithm which takes logarithmic time with respect to the smaller of the two numbers.

The time complexity of the count_common_divisors() function is O(sqrt(gcd(a, b))), as it iterates up to the square root of the gcd of the two numbers.

The space complexity of both functions is O(1), as they only use a constant amount of memory regardless of the input size.