Leetcode 1492: The kth Factor of n

In this Leetcode problem, we should return the kth factor of n, or -1 if such factor doesn’t exist.

Problem statement

Solution

This is pretty much brute force:

• Test all integers from 1 to n
• Count the number of factors current, when current = k return the current factor value

Full code:

class Solution {
    public int kthFactor(int n, int k) {
        int current = 0;
        for (int i = 1; i <= n; i++) {
            if (n % i == 0) {
                current++;
                if (current == k) {
                    return i;
                }
            }
        }
        return -1;
    }
}

Happy coding :)