1492.the-kth-factor-of-n.py

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# @lc app=leetcode id=1492 lang=python3
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# [1492] The kth Factor of n
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# https://leetcode.com/problems/the-kth-factor-of-n/description/
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# algorithms
# Medium (64.52%)
# Likes:    1363
# Dislikes: 259
# Total Accepted:    139.9K
# Total Submissions: 216.5K
# Testcase Example:  '12\n3'
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# 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.
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# Consider a list of all factors of n sorted in ascending order, return the
# k^th factor in this list or return -1 if n has less than k factors.
#
#
# Example 1:
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# Input: n = 12, k = 3
# Output: 3
# Explanation: Factors list is [1, 2, 3, 4, 6, 12], the 3^rd factor is 3.
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# Example 2:
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# Input: n = 7, k = 2
# Output: 7
# Explanation: Factors list is [1, 7], the 2^nd factor is 7.
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# Example 3:
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# Input: n = 4, k = 4
# Output: -1
# Explanation: Factors list is [1, 2, 4], there is only 3 factors. We should
# return -1.
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# Constraints:
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# 1 <= k <= n <= 1000
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# Follow up:
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# Could you solve this problem in less than O(n) complexity?
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#


# @lc code=start
class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        counter = 0
        for i in range(1, n // 2 + 1):
            if n % i == 0:
                counter += 1
            if counter == k:
                return i
        counter += 1
        if counter == k:
            return n
        return -1


# @lc code=end