[TOC] #### Approach #1: Brute Force [Accepted] **Intuition** For each word, check if all prefixes word[:k] are present. We can use a `Set` structure to check this quickly. **Algorithm** Whenever our found word would be superior, we check if all it's prefixes are present, then replace our answer. Alternatively, we could have sorted the words beforehand, so that we know the word we are considering would be the answer if all it's prefixes are present. **Python** ```python class Solution(object): def longestWord(self, words): ans = "" wordset = set(words) for word in words: if len(word) > len(ans) or len(word) == len(ans) and word < ans: if all(word[:k] in wordset for k in xrange(1, len(word))): ans = word return ans ``` *Alternate Implementation* ```python class Solution(object): def longestWord(self, words): wordset = set(words) words.sort(key = lambda c: (-len(c), c)) for word in words: if all(word[:k] in wordset for k in xrange(1, len(word))): return word return "" ``` **Java** ```java class Solution { public String longestWord(String[] words) { String ans = ""; Set wordset = new HashSet(); for (String word: words) wordset.add(word); for (String word: words) { if (word.length() > ans.length() || word.length() == ans.length() && word.compareTo(ans) < 0) { boolean good = true; for (int k = 1; k < word.length(); ++k) { if (!wordset.contains(word.substring(0, k))) { good = false; break; } } if (good) ans = word; } } return ans; } } ``` *Alternate Implementation* ```java class Solution { public String longestWord(String[] words) { Set wordset = new HashSet(); for (String word: words) wordset.add(word); Arrays.sort(words, (a, b) -> a.length() == b.length() ? a.compareTo(b) : b.length() - a.length()); for (String word: words) { boolean good = true; for (int k = 1; k < word.length(); ++k) { if (!wordset.contains(word.substring(0, k))) { good = false; break; } } if (good) return word; } return ""; } } ``` **Complexity Analysis** * Time complexity : $$O(\sum w_i^2)$$, where $$w_i$$ is the length of `words[i]`. Checking whether all prefixes of `words[i]` are in the set is $$O(\sum w_i^2)$$. * Space complexity : $$O(\sum w_i^2)$$ to create the substrings. --- #### Approach #2: Trie + Depth-First Search [Accepted] **Intuition** As prefixes of strings are involved, this is usually a natural fit for a *trie* (a prefix tree.) **Algorithm** Put every word in a trie, then depth-first-search from the start of the trie, only searching nodes that ended a word. Every node found (except the root, which is a special case) then represents a word with all it's prefixes present. We take the best such word. In Python, we showcase a method using defaultdict, while in Java, we stick to a more general object-oriented approach. **Python** ```python class Solution(object): def longestWord(self, words): Trie = lambda: collections.defaultdict(Trie) trie = Trie() END = True for i, word in enumerate(words): reduce(dict.__getitem__, word, trie)[END] = i stack = trie.values() ans = "" while stack: cur = stack.pop() if END in cur: word = words[cur[END]] if len(word) > len(ans) or len(word) == len(ans) and word < ans: ans = word stack.extend([cur[letter] for letter in cur if letter != END]) return ans ``` **Java** ```java class Solution { public String longestWord(String[] words) { Trie trie = new Trie(); int index = 0; for (String word: words) { trie.insert(word, ++index); //indexed by 1 } trie.words = words; return trie.dfs(); } } class Node { char c; HashMap children = new HashMap(); int end; public Node(char c){ this.c = c; } } class Trie { Node root; String[] words; public Trie() { root = new Node('0'); } public void insert(String word, int index) { Node cur = root; for (char c: word.toCharArray()) { cur.children.putIfAbsent(c, new Node(c)); cur = cur.children.get(c); } cur.end = index; } public String dfs() { String ans = ""; Stack stack = new Stack(); stack.push(root); while (!stack.empty()) { Node node = stack.pop(); if (node.end > 0 || node == root) { if (node != root) { String word = words[node.end - 1]; if (word.length() > ans.length() || word.length() == ans.length() && word.compareTo(ans) < 0) { ans = word; } } for (Node nei: node.children.values()) { stack.push(nei); } } } return ans; } } ``` **Complexity Analysis** * Time Complexity: $$O(\sum w_i)$$, where $$w_i$$ is the length of `words[i]`. This is the complexity to build the trie and to search it. If we used a BFS instead of a DFS, and ordered the children in an array, we could drop the need to check whether the candidate word at each node is better than the answer, by forcing that the last node visited will be the best answer. * Space Complexity: $$O(\sum w_i)$$, the space used by our trie.

Longest Word in Dictionary

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Problem Description

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