The kth factor of n

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        count = 0
        for i in range(1, ceil(n/2) + 1):
            if n % i == 0:
                count += 1
                if count == k:
                    return i
        if k == count + 1:
            return n
        return -1

The kth Factor of n

Difficulty: Medium

You are given two positive integers n and k. A factor of an integer n is defined as an integer i where n % i == 0.

Consider a list of all factors of n sorted in ascending order, return the kth factor in this list or return -1 if n has less than k factors.

 

Example 1:

Input: n = 12, k = 3
Output: 3
Explanation: Factors list is [1, 2, 3, 4, 6, 12], the 3rd factor is 3.

Example 2:

Input: n = 7, k = 2
Output: 7
Explanation: Factors list is [1, 7], the 2nd factor is 7.

Example 3:

Input: n = 4, k = 4
Output: -1
Explanation: Factors list is [1, 2, 4], there is only 3 factors. We should return -1.

 

Constraints:

  • 1 <= k <= n <= 1000

 

Follow up:

Could you solve this problem in less than O(n) complexity?