# 1492 The kth Factor of n

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`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 `````` ``````class Solution: def kthFactor(self, n: int, k: int) -> int: divisors, sqrt_n = [], int(n**0.5) for x in range(1, sqrt_n+1): if n % x == 0: k -= 1 if k == 0: return x divisors.append(x) if sqrt_n * sqrt_n == n: k += 1 if k <= len(divisors): return n // divisors[-k] else: return -1``````