花花酱 LeetCode 1362. Closest Divisors

zxi — Sun, 23 Feb 2020 20:00:46 +0000

Given an integer num, find the closest two integers in absolute difference whose product equals num + 1 or num + 2.

Return the two integers in any order.

Example 1:

Input: num = 8
Output: [3,3]
Explanation: For num + 1 = 9, the closest divisors are 3 & 3, for num + 2 = 10, the closest divisors are 2 & 5, hence 3 & 3 is chosen.

Example 2:

Input: num = 123
Output: [5,25]

Example 3:

Input: num = 999
Output: [40,25]

Constraints:

1 <= num <= 10^9

Solution: Brute Force

Time complexity: O(sqrt(n))
Space complexity: O(1)

C++

// Author: Huahua
class Solution {
public:
  vector closestDivisors(int num) {
    auto divisors = [](int n) {
      for (int i = sqrt(n); i >= 2; --i)
        if (n % i == 0) return vector{i, n / i};
      return vector{1, n};
    };
    
    auto ans1 = divisors(num + 1);
    auto ans2 = divisors(num + 2);
    return ans1[1] - ans1[0] < ans2[1] - ans2[0] ? ans1 : ans2;
  }
};

Python3

# Author: Huahua
class Solution:
  def closestDivisors(self, num: int) -> List[int]:
    def divisors(n: int) -> List[int]:
      for i in range(int(math.sqrt(n)), 1, -1):
        if n % i == 0: return [i, n // i]
      return [1, n]
    
    a1, a2 = divisors(num + 1), divisors(num + 2)
    return a1 if a1[1] - a1[0] < a2[1] - a2[0] else a2

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花花酱 LeetCode 1362. Closest Divisors

Solution: Brute Force

C++

Python3