Question

2427. Number of Common Factors

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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 $O(1)$.

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))

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;
  }
};

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);
}

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:
  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

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);
  }
}

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;
  }
};

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)
}

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);
}