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.

• 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)
}
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);
}
}
}

############

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;
}
}

• // 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;
}
};

• class Solution:
def kthFactor(self, n: int, k: int) -> int:
i = 1
while i * i < n:
if n % i == 0:
k -= 1
if k == 0:
return i
i += 1
if i * i != n:
i -= 1
while i:
if (n % (n // i)) == 0:
k -= 1
if k == 0:
return n // i
i -= 1
return -1

• func kthFactor(n int, k int) int {
i := 1
for ; i < n/i; i++ {
if n%i == 0 {
k--
if k == 0 {
return i
}
}
}
if i*i != n {
i--
}
for ; i > 0; i-- {
if n%(n/i) == 0 {
k--
if k == 0 {
return n / i
}
}
}
return -1
}