1415. The k-th Lexicographical String of All Happy Strings of Length n

#Problem

A happy string is a string that:

• consists only of letters of the set ['a', 'b', 'c'].
• s[i] != s[i + 1] for all values of i from 1 to s.length - 1 (string is 1-indexed).

For example, strings “abc”, “ac”, “b” and “abcbabcbcb” are all happy strings and strings “aa”, “baa” and “ababbc” are not happy strings.

Given two integers n and k, consider a list of all happy strings of length n sorted in lexicographical order.

Return the kᵗʰ string of this list or return an empty string if there are less than k happy strings of length n.

https://leetcode.com/problems/the-k-th-lexicographical-string-of-all-happy-strings-of-length-n/

Example 1:

Input: n = 1, k = 3
Output: "c"
Explanation: The list ["a", "b", "c"] contains all happy strings of length 1. The third string is "c".

Example 2:

Input: n = 1, k = 4
Output: ""
Explanation: There are only 3 happy strings of length 1.

Example 3:

Input: n = 3, k = 9
Output: "cab"
Explanation: There are 12 different happy string of length 3 ["aba", "abc", "aca", "acb", "bab", "bac", "bca", "bcb", "cab", "cac", "cba", "cbc"]. You will find the 9ᵗʰ string = "cab"

Constraints:

• 1 <= n <= 10
• 1 <= k <= 100

#Test Cases

class Solution:
    def getHappyString(self, n: int, k: int) -> str:

import pytest

from solution import Solution


@pytest.mark.parametrize('n, k, expected', [
    (1, 3, "c"),
    (1, 4, ""),
    (3, 9, "cab"),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, n, k, expected):
    assert sol.getHappyString(n, k) == expected

#Thoughts

Happy string 除了第一位有 'a'、'b'、'c' 三个选择外，之后的每一位都只有两个选择——即与前一位不同的两个字母。如果用 0 表示这两个字母中较小的那个，用 1 表示较大的那个，显然 happy string 除去第一位之外，剩下的 n - 1 位对应了一个二进制数，一共有 2ⁿ⁻¹ 个可能。而第一位共有三个可能，所以长度为 n 的 happy string 最多有 3 * 2ⁿ⁻¹ 个。

如果 k > 3 * 2ⁿ⁻¹，直接返回空字符串。

否则用 k - 1 除以 2ⁿ⁻¹，商必然在 0 到 2 之间，分别对应 'a'、'b'、'c'；余数转成 n - 1 位的二进制表达，从最高位开始，根据前边对 0 和 1 的定义，就可以逐位构造出 happy string。显然得到的 happy string 就是所有长度为 n 的 happy strings 中按字典序第 k 大的。

时间复杂度 O(n)，空间复杂度 O(n)。

#Code

class Solution:
    def getHappyString(self, n: int, k: int) -> str:
        base = 1 << (n - 1) # 2 ** (n - 1)
        b, k = divmod(k - 1, base)
        if b >= 3:
            return ""

        chars = 'abc'
        candidates = ((1, 2), (0, 2), (0, 1))
        d = b
        parts = [chars[d]]
        while base > 1:
            base >>= 1
            b, k = divmod(k, base)
            d = candidates[d][b]
            parts.append(chars[d])

        return ''.join(parts)