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: 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".
- 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].
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Blog link: https://code.dennyzhang.com/longest-word-in-dictionary ## Basic Ideas: trie tree ## Complexity: Time O(n), Space O(n). n the count of characters involved class TrieNode(object): def __init__(self): self.children = collections.defaultdict(TrieNode) self.is_word = False class Solution(object): def longestWord(self, words): """ :type words: List[str] :rtype: str """ # Build TrieNode root = TrieNode() # check each word, and insert if missing for word in words: # always check from the top node = root for ch in word: node = node.children[ch] node.is_word = True return self.foundLongestWord(root) def foundLongestWord(self, node): """ :rtype: (length, str) """ # BFS: # How to check: # Candidates should be: # 1. is_word as true for all nodes in the path. # 2. Has no children # How to move to next: # Only check nodes with is_word as True # When node has no children, we max_length, max_str = 0, '' queue =  # initialize queue # Since we have sorted the keys, we will get smallest lexicographical match for ch in sorted(node.children): child = node.children[ch] if child.is_word: queue.append((child, ch, 1)) while len(queue) != 0: (node, str, length) = queue del queue if length > max_length: max_length, max_str = length, str for ch in sorted(node.children): child = node.children[ch] if child.is_word: queue.append((child, '%s%s' % (str, ch), length+1)) return max_str