LeetCode-in-Java

720. Longest Word in Dictionary

Medium

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.

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.

Solution

public class Solution {
    private final Node root = new Node();
    private String longestWord = "";

    private static class Node {
        Node[] children;
        String str;

        public Node() {
            this.children = new Node[26];
            this.str = null;
        }

        public void insert(Node curr, String word) {
            for (int i = 0; i < word.length(); i++) {
                char ch = word.charAt(i);
                if (curr.children[ch - 'a'] == null) {
                    curr.children[ch - 'a'] = new Node();
                }
                curr = curr.children[ch - 'a'];
            }
            curr.str = word;
        }
    }

    public String longestWord(String[] words) {
        for (String word : words) {
            root.insert(root, word);
        }
        dfs(root);
        return longestWord;
    }

    private void dfs(Node curr) {
        for (int i = 0; i < 26; i++) {
            if (curr.children[i] != null && curr.children[i].str != null) {
                if (curr.children[i].str.length() > longestWord.length()) {
                    longestWord = curr.children[i].str;
                }
                dfs(curr.children[i]);
            }
        }
    }
}

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