Permutation Sequence

The set [1,2,3,...,*n*] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order, we get the following sequence for n = 3:

• "123"- "132"- "213"- "231"- "312"- "321"

Given n and k, return the k^th permutation sequence.

Note:

• Given n will be between 1 and 9 inclusive.- Given k will be between 1 and n! inclusive.

Example 1:

**Input:** n = 3, k = 3
**Output:** "213"

Example 2:

**Input:** n = 4, k = 9
**Output:** "2314"

class Solution:
    def getPermutation(self, n: int, k: int) -> str:
        res, nums = "", list(range(1, n + 1))

        k -= 1

        for n in range(n, 0, -1):
            index, k = divmod(k, math.factorial(n-1))
            res += str(nums.pop(index))

        return res