60. Permutation Sequence

Problem:

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:

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the kth 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"

Solution:

The order of the sequence is fixed hence can be calculated. We can view the process as picking digits from a sorted set [1...n].

Each digit appears (n-1)! times in result[0]. And for a fixed result[0] each digit appears (n-2)! times in result[1]. So on.

We also need k-- to convert k into index so that k <= (n-1)! maps 0 (and get 1 from the set).

/**
 * @param {number} n
 * @param {number} k
 * @return {string}
 */
var getPermutation = function(n, k) {
  const digits = []
  let factorial = 1
  for (let i = 1; i <= n; i++) {
    digits.push(i)
    factorial *= i
  }

  k--

  let result = ''
  while (n > 0) {
    factorial /= n
    result += digits.splice(k / factorial | 0, 1)[0]
    k %= factorial
    n--
  }

  return result
};

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