# 2427. Number of Common Factors

## Description

Given two positive integers a and b, return the number of common factors of a and b.

An integer x is a common factor of a and b if x divides both a and b.

Example 1:

Input: a = 12, b = 6
Output: 4
Explanation: The common factors of 12 and 6 are 1, 2, 3, 6.


Example 2:

Input: a = 25, b = 30
Output: 2
Explanation: The common factors of 25 and 30 are 1, 5.


Constraints:

• 1 <= a, b <= 1000

## Solutions

Solution 1: Enumeration

We can first calculate the greatest common divisor $g$ of $a$ and $b$, then enumerate each number in $[1,..g]$, check whether it is a factor of $g$, if it is, then increment the answer by one.

The time complexity is $O(\min(a, b))$, and the space complexity is $O(1)$.

Solution 2: Optimized Enumeration

Similar to Solution 1, we can first calculate the greatest common divisor $g$ of $a$ and $b$, then enumerate all factors of the greatest common divisor $g$, and accumulate the answer.

The time complexity is $O(\sqrt{\min(a, b)})$, and the space complexity is $O(1)$.

• class Solution {
public int commonFactors(int a, int b) {
int g = gcd(a, b);
int ans = 0;
for (int x = 1; x <= g; ++x) {
if (g % x == 0) {
++ans;
}
}
return ans;
}

private int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
}

• class Solution {
public:
int commonFactors(int a, int b) {
int g = gcd(a, b);
int ans = 0;
for (int x = 1; x <= g; ++x) {
ans += g % x == 0;
}
return ans;
}
};

• class Solution:
def commonFactors(self, a: int, b: int) -> int:
g = gcd(a, b)
return sum(g % x == 0 for x in range(1, g + 1))


• func commonFactors(a int, b int) (ans int) {
g := gcd(a, b)
for x := 1; x <= g; x++ {
if g%x == 0 {
ans++
}
}
return
}

func gcd(a int, b int) int {
if b == 0 {
return a
}
return gcd(b, a%b)
}

• function commonFactors(a: number, b: number): number {
const g = gcd(a, b);
let ans = 0;
for (let x = 1; x <= g; ++x) {
if (g % x === 0) {
++ans;
}
}
return ans;
}

function gcd(a: number, b: number): number {
return b === 0 ? a : gcd(b, a % b);
}