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Tested Python solution for LeetCode 60 with 16 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 60 # by problem number
lcpy gen -s permutation_sequence # by problem name
[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"n and k, return the kth permutation sequence.
Input: Input: n = 3, k = 3
Output: "213"
Input: Input: n = 4, k = 9
Output: "2314"
Input: Input: n = 3, k = 1
Output: "123"
1 <= n <= 91 <= k <= n!import math
class Solution:
# Time: O(n^2)
# Space: O(n)
def get_permutation(self, n: int, k: int) -> str:
digits = [str(i) for i in range(1, n + 1)]
remaining = k - 1
parts: list[str] = []
for i in range(n, 0, -1):
block_size = math.factorial(i - 1)
idx, remaining = divmod(remaining, block_size)
parts.append(digits.pop(idx))
return "".join(parts)
| Time | Space |
|---|---|
| O(n^2) | O(n) |