Implement Trie (Prefix Tree)

Problem

Implement a trie with insert, search, and startsWith methods.

Note: You may assume that all inputs are consist of lowercase letters a-z.

Solution

class Trie {

    private class TrieNode {
        TrieNode[] children;
        boolean endOfWord;

        public TrieNode() {
            this.children = new TrieNode[26];
            this.endOfWord = false;
        }
    }

    private TrieNode root;
    /** Initialize your data structure here. */
    public Trie() {
        this.root = new TrieNode();
    }

    /** Inserts a word into the trie. */
    public void insert(String word) {
        TrieNode cur = root;
        for (char ch : word.toCharArray()) {
            if (cur.children[ch - 'a'] == null) cur.children[ch - 'a'] = new TrieNode();
            cur = cur.children[ch - 'a'];
        }
        cur.endOfWord = true;
    }

    /** Returns if the word is in the trie. */
    public boolean search(String word) {
        TrieNode cur = root;
        for (char ch : word.toCharArray()) {
            if (cur.children[ch - 'a'] == null) return false;
            cur = cur.children[ch - 'a'];
        }
        return cur.endOfWord;
    }

    /** Returns if there is any word in the trie that starts with the given prefix. */
    public boolean startsWith(String prefix) {
        TrieNode cur = root;
        for (char ch : prefix.toCharArray()) {
            if (cur.children[ch - 'a'] == null) return false;
            cur = cur.children[ch - 'a'];
        }
        return true;
    }
}

/**
 * Your Trie object will be instantiated and called as such:
 * Trie obj = new Trie();
 * obj.insert(word);
 * boolean param_2 = obj.search(word);
 * boolean param_3 = obj.startsWith(prefix);
 */

Analysis

This is the basic implementation of Trie data structure
Generally, every Trie object has a Map<Character, Trie> as children and endOfWord to indicate if it's a word
Each letter from a word is connected by the map
In this problem, since we only have 26 letters at most, we use Trie[] to replace the map
children[ch - 'a'] <=> map.get(ch)