LeetCode - #60 Permutation Sequence

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Difficulty Level: Difficult

1. Description

Gives a set [1,2,3,...,n]with all elements in a total n!of permutations.

List all arrangements in order of magnitude and mark them one by one. n = 3When , all arrangements are as follows:

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

Given nand k, return kthe permutation.

2. Example

Example 1

输入：n = 3, k = 3
输出："213"
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Example 2

输入：n = 4, k = 9
输出："2314"
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Example 3

输入：n = 3, k = 1
输出："123"
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Restrictions:

1 <= n <= 9
1 <= k <= n!

3. Answers

class PermutationSequence {
    func getPermutation(_ n: Int, _ k: Int) -> String {
        var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9]

        var factorial = 1
        for i in 1 ..< n {
            factorial *= i
        }

        var result = ""
        var k = k
        var divisor = n - 1

        for i in 0 ..< n {
            for (index, number) in numbers.enumerated() {
                if k > factorial {
                    k -= factorial
                } else {
                    result += "\(number)"
                    numbers.remove(at: index)
                    break
                }
            }
            if divisor > 1 {
                factorial /= divisor
                divisor -= 1
            }
        }

        return result
    }
}
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Main idea: iterate and change the array from last to first.
Time Complexity: O(n^2)
Space Complexity: O(1)

Repository for the algorithm solution: LeetCode-Swift

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