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 [1], 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)
                    factors.add(i);
            }
            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);
            }
        }
    }
    
  • // 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;
        }
    };
    
  • print("Todo!")
    

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