Formatted question description: https://leetcode.ca/all/1492.html

# 1492. The kth Factor of n (Medium)

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 , 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

Related Topics:
Math

## Solution 1.

// OJ: https://leetcode.com/problems/the-kth-factor-of-n/

// Time: O(sqrt(N))
// Space: O(1)
class Solution {
public:
int kthFactor(int n, int k) {
int i = 1;
for (; i * i <= n; ++i) {
if (n % i) continue;
if (--k == 0) return i;
}
--i;
if (i * i == n) --i;
for (; i >= 1; --i) {
if (n % i) continue;
if (--k == 0) return n / i;
}
return -1;
}
};


Java

class Solution {
public int kthFactor(int n, int k) {
if (k == 1)
return 1;
int sqrt = (int) Math.sqrt(n);
List<Integer> factors = new ArrayList<Integer>();
for (int i = 1; i <= sqrt; i++) {
if (n % i == 0)
}
int size = factors.size();
int factorsCount = sqrt * sqrt == n ? size * 2 - 1 : size * 2;
if (k > factorsCount)
return -1;
else if (k <= size)
return factors.get(k - 1);
else {
int index = factorsCount - k;
return n / factors.get(index);
}
}
}