# 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

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
}

• function kthFactor(n: number, k: number): number {
for (let i = 1; i <= n; ++i) {
if (n % i === 0 && --k === 0) {
return i;
}
}
return -1;
}