1585. The Kth Factor Of N¶

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1585. The kth Factor of n

Medium

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¶

class Solution {
    public int kthFactor(int n, int k) {
        ArrayList<Integer> factors = new ArrayList<>();
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                factors.add(i);
                if (n / i != i) factors.add(n / i);
            }
        }
        Collections.sort(factors);
        if (factors.size() < k) return -1;
        return factors.get(k - 1);
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here