2.py

# 1492. The kth Factor of n
# User Accepted:4524
# User Tried:4554
# Total Accepted:4600
# Total Submissions:6166
# Difficulty:Medium
# 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.

# Example 4:
# Input: n = 1, k = 1
# Output: 1
# Explanation: Factors list is [1], the 1st factor is 1.

# Example 5:
# Input: n = 1000, k = 3
# Output: 4
# Explanation: Factors list is [1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000].
 
# Constraints:
# # 1 <= k <= n <= 1000

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        res = []
        for i in range(1, n+1):
            if n % i == 0:
                res.append(i)
        if k > len(res):
            return -1
        return res[k-1]