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?
Solutions
Solution 1: Brute Force Enumeration
A "factor" is a number that can divide another number. Therefore, we only need to enumerate from 1 to n, find all numbers that can divide n, and then return the k-th one.
The time complexity is O(n), and the space complexity is O(1).
class Solution: def kthFactor(self, n: int, k: int) -> int: for i in range(1, n + 1): if n % i == 0: k -= 1 if k == 0: return i return -1(code-box)
Solution 2: Optimized Enumeration
We can observe that if n has a factor x, then n must also have a factor n/x.
Therefore, we first need to enumerate [1,2,...\left \lfloor √n \right \rfloor], find all numbers that can divide n. If we find the k-th factor, then we can return it directly. If we do not find the k-th factor, then we need to enumerate [\left \lfloor √n \right \rfloor ,..1] in reverse order, and find the k-th factor.
The time complexity is O(√n), and the space complexity is O(1).
class Solution: def kthFactor(self, n: int, k: int) -> int: i = 1 while i * i < n: if n % i == 0: k -= 1 if k == 0: return i i += 1 if i * i != n: i -= 1 while i: if (n % (n // i)) == 0: k -= 1 if k == 0: return n // i i -= 1 return -1(code-box)
