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".

Note:

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].

Analysis

See: https://leetcode.com/problems/longest-word-in-dictionary/solution/

Trie + Depth-First Search

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.

Solution

class Solution {
    public String longestWord(String[] words) {
        Trie trie = new Trie();
        for (String word : words) {
             trie.insert(word);
        }
        return dfs(trie.getRoot());
    }
    String dfs(TrieNode root) {
         String longest = "";
         Stack<TrieNode> stack = new Stack();
         stack.push(root);
         while (!stack.isEmpty()) {
             TrieNode node = stack.pop();
             if (node == null) continue;
             if (node.isEnd == true || node == root) {
                 if (node != root) {
                     if (node.word.length() > longest.length() || 
                         (node.word.length() == longest.length() && node.word.compareTo(longest) < 0)) {
                         longest = node.word;
                     }
                 }
                 for (TrieNode child: node.children) {
                    stack.push(child);
                 }
             }
         }
         return longest;
    }
}

class Trie {
    TrieNode root;
    public Trie() {
        root = new TrieNode();
    }
    public void insert(String word) {
        TrieNode node = root;
        char[] charArray = word.toCharArray();
        for (char ch : charArray) {
            if (!node.contains(ch)) {
                node.put(ch, new TrieNode());
            }
            node = node.get(ch);
        }
        node.isEnd = true;
        node.word = word;
    }

    public TrieNode getRoot() {
        return root;
    }
}
class TrieNode {
    TrieNode[] children;
    boolean isEnd;
    String word; 
    public TrieNode () {
        children = new TrieNode[26];
    }
    boolean contains(char ch) {
        return children[ch - 'a'] != null;
    }
    TrieNode get(char ch) {
        return children[ch - 'a'];
    }
    void put(char ch, TrieNode node) {
        children[ch - 'a'] = node;
    }
}