Longest Word in Dictionary Using Trie in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Trie Insertion and Search

What is the output of the following code that inserts words into a Trie and searches for a prefix?

DSA Javascript

class TrieNode {
  constructor() {
    this.children = {};
    this.isEnd = false;
  }
}

class Trie {
  constructor() {
    this.root = new TrieNode();
  }

  insert(word) {
    let node = this.root;
    for (const char of word) {
      if (!node.children[char]) {
        node.children[char] = new TrieNode();
      }
      node = node.children[char];
    }
    node.isEnd = true;
  }

  startsWith(prefix) {
    let node = this.root;
    for (const char of prefix) {
      if (!node.children[char]) {
        return false;
      }
      node = node.children[char];
    }
    return true;
  }
}

const trie = new Trie();
trie.insert("apple");
trie.insert("app");
console.log(trie.startsWith("app"));
console.log(trie.startsWith("apl"));

A true\nfalse

B true\ntrue

C false\nfalse

D false\ntrue

💡 Hint

Think about whether the prefix 'app' exists in the Trie after insertion.

The prefix 'app' exists because both 'app' and 'apple' were inserted. The prefix 'apl' does not exist, so startsWith returns false for it.

🧠 Conceptual

intermediate

Longest Word Construction Condition

In the problem of finding the longest word in a dictionary using a Trie, what condition must be true for a word to be considered valid for the longest word?

A The word must not share any prefix with other words

B The word must be the shortest in the dictionary

C Every prefix of the word must also be a word in the dictionary

D The word must contain at least one vowel

💡 Hint

Think about building the word step by step from smaller words.

The longest word is valid only if all its prefixes are also words in the dictionary, ensuring it can be built one character at a time.

🔧 Debug

advanced

Identify the Bug in Trie Longest Word Search

What error does the following code produce when trying to find the longest word in a Trie? class TrieNode { constructor() { this.children = {}; this.isEnd = false; } } class Trie { constructor() { this.root = new TrieNode(); } insert(word) { let node = this.root; for (const char of word) { if (!node.children[char]) { node.children[char] = new TrieNode(); } node = node.children[char]; } node.isEnd = true; } longestWord() { let result = ""; const dfs = (node, word) => { if (!node.isEnd && word.length !== 0) return; if (word.length > result.length || (word.length === result.length && word < result)) { result = word; } for (const char in node.children) { dfs(node.children[char], word + char); } }; dfs(this.root, ""); return result; } } const trie = new Trie(); ["w","wo","wor","worl","world"].forEach(w => trie.insert(w)); console.log(trie.longestWord());

A TypeError: Cannot read property 'isEnd' of undefined

B SyntaxError: Unexpected token

C Returns empty string ""

D Returns "world" correctly

💡 Hint

Check the base case and traversal logic in DFS.

The code correctly returns 'world' as the longest word because all prefixes are inserted and isEnd is checked properly.

🚀 Application

advanced

Longest Word Result After Insertions

Given the following words inserted into a Trie: ["a", "banana", "app", "appl", "ap", "apply", "apple"], what is the longest word returned by the longestWord() method that requires all prefixes to be valid words?

A "apple"

B "appl"

C "apply"

D "banana"

💡 Hint

Check which words have all prefixes present in the list.

Both "apple" and "apply" have all prefixes present, but "apple" comes before "apply" lexicographically, so it is returned.

🧠 Conceptual

expert

Why Use Trie for Longest Word in Dictionary?

Why is a Trie data structure preferred over sorting and checking prefixes directly when finding the longest word in a dictionary where all prefixes must be valid words?

A Sorting is always slower than Trie insertion

B Trie allows efficient prefix checking and incremental building of words in O(N * L) time where N is number of words and L is max length

C Trie uses less memory than arrays or hash maps

D Trie automatically sorts words lexicographically without extra code

💡 Hint

Consider how prefix checks are done repeatedly.

Trie enables fast prefix checks and incremental word building, making it efficient for this problem compared to repeated prefix checks after sorting.