[LeetCode] 720. Longest Word in Dictionary

- Problem

720. Longest Word in Dictionary
Given an array of strings words representing an English Dictionary, return the longest word in words that can be built one character at a time by other words in words.

If there is more than one possible answer, return the longest word with the smallest lexicographical order. If there is no answer, return the empty string.

Note that the word should be built from left to right with each additional character being added to the end of a previous word.

Example 1:

Input: words = ["w","wo","wor","worl","world"]
Output: "world"
Explanation: The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".

Example 2:

Input: words = ["a","banana","app","appl","ap","apply","apple"]
Output: "apple"
Explanation: Both "apply" and "apple" can be built from other words in the dictionary. However, "apple" is lexicographically smaller than "apply".

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length <= 30
• words[i] consists of lowercase English letters.

- 내 풀이

class TrieNode:
    def __init__(self):
        self.children = dict()
        self.is_word = False
        self.word = ""

class Trie:
    def __init__(self):
        self.root = TrieNode()
        
    def add(self, word):
        curr = self.root
        
        for c in word:
            if c not in curr.children:
                curr.children[c] = TrieNode()
            curr = curr.children[c]
        
        curr.is_word = True
        curr.word = word
    
    def bfs(self):
        q = collections.deque([self.root])
        answer = []
        
        while q:
            curr = q.popleft()
            for i in curr.children.values():
                if i.is_word:
                    q.append(i)
                    answer.append(i.word)
        
        return answer
        
class Solution:
    def longestWord(self, words: List[str]) -> str:
        trie = Trie()
        
        for word in words:
            trie.add(word)
        
        answer = trie.bfs()
        
        return sorted(answer, key = lambda x : (-len(x), x))[0] if answer else ""

- 결과