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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 $O(1)$.

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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))

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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);
    }
}

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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;
    }
};

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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)
}

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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);
}

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 $O(1)$.

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class Solution:
    def commonFactors(self, a: int, b: int) -> int:
        g = gcd(a, b)
        ans, x = 0, 1
        while x * x <= g:
            if g % x == 0:
                ans += 1
                ans += x * x < g
            x += 1
        return ans

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

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

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

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

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

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

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

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