# 720. Longest Word in Dictionary

Given a list of strings words representing an English Dictionary, find 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.

Example 1:
words = ["w","wo","wor","worl", "world"]
Output: "world"
The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".

Example 2:
words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]
Output: "apple"
Both "apply" and "apple" can be built from other words in the dictionary. However, "apple" is lexicographically smaller than "apply".


All the strings in the input will only contain lowercase letters.
The length of words will be in the range [1, 1000].
The length of words[i] will be in the range [1, 30].

# Solution

Approach 1: use a set to speed up all look ups. Can presort the word list to speed up prefix checks. Time: O(N).

Approach 2: use a trie to keep all words, and later go through either the trie or the word again. Can also presort the word list to reduce the number of times needed to look down the trie. Time: O(NL).

# Code (Python)

Approach 1:

    def longestWord(self, words: 'List[str]') -> 'str':
        words = sorted(words)
        words_can_be_built = set()
        result = ''
        for word in words:
            if len(word) == 1 or word[:-1] in words_can_be_built:
                if len(word) > len(result) or (len(word) == len(result) and word < result):
                    result = word
        return result

Approach 2:

class Solution:
    def longestWord(self, words: List[str]) -> str:
        words.sort(key=lambda word: (len(word), word))
        root = TrieNode(is_word=True)
        longest_word = ''
        for word in words:
            # insert into Trie, validing is_word along the way
            skip = False
            node = root
            for letter in word:
                if not node.is_word:
                    skip = True
                if letter not in node.children:
                    node.children[letter] = TrieNode()
                node = node.children[letter]
            if not skip:
                node.is_word = True
                if len(word) > len(longest_word):
                    longest_word = word
        return longest_word

class TrieNode:
    def __init__(self, is_word=False):
        self.is_word = is_word
        self.children = {}

# Code (C++)

Approach 1:

Approach 2: