Welcome to Subscribe On Youtube
1492. The kth Factor of n
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
-
class Solution { public int kthFactor(int n, int k) { for (int i = 1; i <= n; ++i) { if (n % i == 0 && (--k == 0)) { return i; } } return -1; } } -
class Solution { public: int kthFactor(int n, int k) { int i = 1; for (; i < n / i; ++i) { if (n % i == 0 && (--k == 0)) { return i; } } if (i * i != n) { --i; } for (; i > 0; --i) { if (n % (n / i) == 0 && (--k == 0)) { return n / i; } } return -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 -
func kthFactor(n int, k int) int { for i := 1; i <= n; i++ { if n%i == 0 { k-- if k == 0 { return i } } } return -1 }