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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 amap ma; // Function that calculate the count of// each prime factor of a numbervoid 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 numbersint 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 programimport 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 programimport math # Map to store the count of each# prime factor of ama = {} # Function that calculate the count of# each prime factor of a numberdef 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 numbersdef 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 = 12b = 24 print(commDiv(a, b)) # This code is contributed by Stream_Cipher

## C#

 // C# implementation of programusing System;using System.Collections.Generic; class GFG{   // Map to store the count of each// prime factor of astatic Dictionary ma = new Dictionary(); // Function that calculate the count of// each prime factor of a numberstatic 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 numbersstatic 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 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 numbersint 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 numbersint 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 caseint 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 programfrom math import sqrt  # Function to calculate gcd of two numbersdef 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 numbersdef 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 caseif __name__ == "__main__":    a = 12    b = 24;    print(commDiv(a, b))

## C#

 // C# implementation of programusing 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 numbersfunction 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 = 12b = 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 algorithmfunction 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 bfunction 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.