Problem
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?
Solution (Java)
class Solution {
public int kthFactor(int n, int k) {
List<Integer> list = new ArrayList<>();
for (int i = 1; i <= n; i++) {
if (n % i == 0) {
list.add(i);
}
}
return list.size() >= k ? list.get(k - 1) : -1;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).