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

Difficulty: interview

Starter code:

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        

Solutions:
class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        i = 0
        for j in range(1, n+1):
            if n % j == 0:
                i += 1
                if i == k:
                    return j
        return -1
            


class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        r1 = []
        r2 = []
        for i in range(1, int(n**.5) + 1):
            if n % i == 0:
                r1.append(i)
                if n != i * i:
                    r2 = [n//i] + r2
        r = r1 + r2
        if len(r) < k:
            return -1
        else:
            return r[k - 1]

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        i = 1
        # factors that are smaller than sqrt(n)
        factor_list_small = []
        # factors that are larger than sqrt(n)
        factor_list_large = []
        while pow(i,2) < n:
            a, b = divmod(n, i) # a is the quotient, b is the remainder
            if b == 0:
                factor_list_small.append(i) # add to the end
                factor_list_large = [a] + factor_list_large # add to the beginning 
            i += 1
        if pow(i,2) == n:
            factor_list_small.append(i)
        # combine these two lists together
        factor_list = factor_list_small + factor_list_large
        if len(factor_list) < k:
            return(-1)
        else:
            return(factor_list[k-1])

from math import sqrt, ceil

class Solution:
    def kthFactor(self, n, k):
        factors = []
        if int(sqrt(n)) == sqrt(n):
            for i in range(1, int(sqrt(n))+1):
                if n % i == 0:
                    factors.append(i)
                    if i == int(n/i):
                        continue
                    factors.append(int(n / i))
        else:
            for i in range(1, ceil(sqrt(n))):
                if n % i == 0:
                    factors.append(i)
                    factors.append(int(n/i))
        factors.sort()
        if k > len(factors):
            return -1
        return factors[k-1]

import math
class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        facts=[]
        for i in range(1,n+1):
            if n%i==0:
                facts.append(i)
        if len(facts)<k:
            return -1
        else:
            return facts[k-1]
            


class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        check = []
        for i in range(1,n+1):
            if n % i == 0:
                check.append(i)
        if len(check) < k:
            return -1
        else:
            return check[k-1]

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        i = 0
        for j in range(1, n + 1):
            if n%j == 0:
                i += 1
            if i == k:
                return j
        return -1


class Solution:
    def kthFactor(self, n, k):
        
        solution_space = list([x for x in list(range(1,n+1)) if n % x == 0])
        if len(solution_space) < k:
            return -1
        else:
            return list([x for x in list(range(1,n+1)) if n % x == 0])[k-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

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        try:
            return list(filter(lambda x:n%x==0,range(1,n+1)))[k-1]
        except:
            return -1