[leetCode] 1492. The kth Factor of n (Python)

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

Solution

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        loop_num = n // 2
        
        if n == 1:
            divisors = [1]
        else:
            divisors = []

            for i in range(1, loop_num+1):
                if n % i == 0:
                    divisors.append(i)
                    divisors.append(n // i)

            divisors = sorted(list(set(divisors)))
        
        
        if len(divisors) < k:
            return -1
        
        return divisors[k-1]

SOMJANG/CODINGTEST_PRACTICE