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1492. The kth Factor of n

Description

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?

Solutions

  • class Solution {
        public int kthFactor(int n, int k) {
            for (int i = 1; i <= n; ++i) {
                if (n % i == 0 && (--k == 0)) {
                    return i;
                }
            }
            return -1;
        }
    }
    
  • class Solution {
    public:
        int kthFactor(int n, int k) {
            int i = 1;
            for (; i < n / i; ++i) {
                if (n % i == 0 && (--k == 0)) {
                    return i;
                }
            }
            if (i * i != n) {
                --i;
            }
            for (; i > 0; --i) {
                if (n % (n / i) == 0 && (--k == 0)) {
                    return n / i;
                }
            }
            return -1;
        }
    };
    
  • class Solution:
        def kthFactor(self, n: int, k: int) -> int:
            for i in range(1, n + 1):
                if n % i == 0:
                    k -= 1
                    if k == 0:
                        return i
            return -1
    
    
  • func kthFactor(n int, k int) int {
    	for i := 1; i <= n; i++ {
    		if n%i == 0 {
    			k--
    			if k == 0 {
    				return i
    			}
    		}
    	}
    	return -1
    }
    
  • function kthFactor(n: number, k: number): number {
        for (let i = 1; i <= n; ++i) {
            if (n % i === 0 && --k === 0) {
                return i;
            }
        }
        return -1;
    }
    
    

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