Permutation sequence

Time: O(N^2); Space: O(N); medium

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 (i.e., for n = 3):

"123"
"132"
"213"
"231"
"312"
"321"

Given n and k, return the kth permutation sequence.

Example 1:

Input: n = 3, k = 3

Output: “213”

Example 2:

Input: n = 4, k = 9

Output: “2314”

Notes:

  • Given n will be between 1 and 9 inclusive.

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

1. Cantor ordering solution

[1]:

import math

class Solution1(object):
    def getPermutation(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: str
        """
        seq, k, fact = "", k - 1, math.factorial(n - 1)
        perm = [i for i in range(1, n + 1)]

        for i in reversed(range(n)):
            curr = perm[k // fact]
            seq += str(curr)
            perm.remove(curr)
            if i > 0:
                k %= fact
                fact //= i
        return seq

[3]:

s = Solution1()
n = 3
k = 3
assert s.getPermutation(n, k) == "213"
n = 4
k = 9
assert s.getPermutation(n, k) == "2314"