LeetCode in Kotlin

60. Permutation Sequence

Hard

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.

Example 1:

Input: n = 3, k = 3

Output: “213”

Example 2:

Input: n = 4, k = 9

Output: “2314”

Example 3:

Input: n = 3, k = 1

Output: “123”

Constraints:

• 1 <= n <= 9
• 1 <= k <= n!

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    fun getPermutation(n: Int, k: Int): String {
        var k = k
        val res = CharArray(n)
        // We want the permutation sequence to be zero-indexed
        k = k - 1
        // The set bits indicate the available digits
        var a = (1 shl n) - 1
        var m = 1
        for (i in 2 until n) {
            // m = (n - 1)!
            m *= i
        }
        for (i in 0 until n) {
            var b = a
            for (j in 0 until k / m) {
                b = b and b - 1
            }
            // b is the bit corresponding to the digit we want
            b = b and b.inv() + 1
            res[i] = ('1'.code + Integer.bitCount(b - 1)).toChar()
            // Remove b from the set of available digits
            a = a and b.inv()
            k %= m
            m /= Math.max(1, n - i - 1)
        }
        return String(res)
    }
}

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