leetcode1415. The k-th Lexicographical String of All Happy Strings of Length n

Posted 2020-12-06 seyjs

题目如下：

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.

 

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"

Example 4:

Input: n = 2, k = 7
Output: ""

Example 5:

Input: n = 10, k = 100
Output: "abacbabacb"

Constraints:

• 1 <= n <= 10
• 1 <= k <= 100

解题思路：n和k都很小，最简单的方法是把所有符合条件的string都求出来，排序后可得到结果。

代码如下：

class Solution(object):
    def getHappyString(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: str
        """
        queue = [‘a‘,‘b‘,‘c‘]

        l = []
        while len(queue) > 0:
            item = queue.pop(0)
            if len(item) == n:
                l.append(item)
                continue
            if item[-1] == ‘a‘:
                queue.append(item + ‘b‘)
                queue.append(item + ‘c‘)
            elif item[-1] == ‘b‘:
                queue.append(item + ‘a‘)
                queue.append(item + ‘c‘)
            elif item[-1] == ‘c‘:
                queue.append(item + ‘a‘)
                queue.append(item + ‘b‘)

        l.sort()
        if len(l) < k:
            return ‘‘
        return l[k-1]