Description
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?
Solution
Python3
class Solution:
def kthFactor(self, n: int, k: int) -> int:
res = set()
for i in range(1, n+1):
if n % i == 0:
res.add(i)
res.add(n//i)
res = sorted(list(res))
return res[k-1] if len(res) >= k else -1