Longest Word in Dictionary Using Trie in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Trie data structure?

A Trie is a tree-like data structure used to store a dynamic set of strings where keys are usually strings. Each node represents a character, and paths from root to nodes represent words or prefixes.

intermediate

How does a Trie help in finding the longest word in a dictionary?

A Trie allows efficient prefix checking. By building a Trie from dictionary words, we can traverse nodes only if the prefix is a valid word, helping find the longest word where all prefixes exist.

intermediate

What is the main condition to continue traversing in the Trie when searching for the longest word?

We only continue traversing to child nodes if the current node marks the end of a valid word, ensuring all prefixes of the candidate word exist in the dictionary.

intermediate

Why do we prefer DFS (Depth-First Search) over BFS (Breadth-First Search) in finding the longest word in a Trie?

DFS explores deeper paths first, which helps find longer words by going down the Trie branches fully. BFS finds shorter words first, which is less efficient for this problem.

beginner

What is the time complexity of building a Trie from n words with average length m?

The time complexity is O(n * m) because each word of length m is inserted character by character into the Trie.

What does each node in a Trie represent?

A A number

B A whole word

C A character of a word

D A sentence

Each node in a Trie represents a single character of a word.

When searching for the longest word in a Trie, you should only continue if:

A The current node is a leaf

B The current node has any children

C The current node is the root

D The current node is the end of a valid word

We continue only if the current node marks the end of a valid word to ensure all prefixes exist.

Which traversal method is best to find the longest word in a Trie?

A Inorder Traversal

B Depth-First Search (DFS)

C Level Order Traversal

D Breadth-First Search (BFS)

DFS explores deeper paths first, helping find the longest word.

What is the main advantage of using a Trie for prefix-based problems?

A Efficient prefix checking

B Sorting words alphabetically

C Storing numbers

D Reducing memory usage

Trie allows efficient checking if a prefix exists in the dictionary.

If you have 1000 words each of length 10, what is the approximate time to build a Trie?

A O(10,000)

B O(1,000)

C O(10)

D O(100)

Building a Trie takes O(n * m), here 1000 * 10 = 10,000 operations.

Explain how a Trie is used to find the longest word in a dictionary where every prefix is also a word.

Think about checking prefixes and exploring deep paths.

Describe the steps to build a Trie and then find the longest word with all prefixes present.

Focus on insertion and traversal conditions.

Practice

1. What is the main purpose of using a Trie data structure when finding the longest word in a dictionary?

easy

A. To reverse the words in the dictionary

B. To sort words alphabetically

C. To count the total number of words

D. To efficiently store and search words by their prefixes

Solution

1. Step 1: Understand Trie usage

   A Trie stores words by breaking them into characters and linking nodes for prefixes.

2. Step 2: Connect Trie to longest word search

   This structure allows quick prefix checks, which helps find the longest word built from other words.

3. Final Answer:

   To efficiently store and search words by their prefixes -> Option D

4. Quick Check:

   Trie = prefix search [OK]

Hint: Trie is all about prefixes and fast word lookup [OK]

Common Mistakes:

• Confusing Trie with sorting algorithms
• Thinking Trie reverses words
• Assuming Trie counts words only

2. Which of the following is the correct way to insert a word into a Trie node in JavaScript?

easy

A. node[word[i]] = new TrieNode() if not exists

B. node[word[i]] = word[i]

C. node.push(word[i])

D. node[word] = true

Solution

1. Step 1: Understand Trie node insertion

   Each character of the word is a key in the node object; if missing, create a new TrieNode.

2. Step 2: Identify correct syntax

   Using node[word[i]] = new TrieNode() creates a child node for that character if it doesn't exist.

3. Final Answer:

   node[word[i]] = new TrieNode() if not exists -> Option A

4. Quick Check:

   Insert char as child node [OK]

Hint: Insert chars as keys with new nodes if missing [OK]

Common Mistakes:

• Assigning character directly instead of node
• Using array push on object
• Setting whole word as key instead of chars

3. Given the words ["a", "app", "apple", "apply", "ap"], what is the longest word that can be built one character at a time using the Trie method?

medium

A. "apple"

B. "apply"

C. "app"

D. "ap"

Solution

1. Step 1: Check each word buildability

   Words must be built one char at a time with all prefixes present. "a", "ap", "app" are present, but "apple" and "apply" require "appl" which is missing.

2. Step 2: Identify longest valid word

   "app" is the longest word where all prefixes ("a", "ap", "app") exist in the list.

3. Final Answer:

   "app" -> Option C

4. Quick Check:

   Longest buildable word = "app" [OK]

Hint: Check all prefixes exist for each candidate word [OK]

Common Mistakes:

• Assuming "apple" or "apply" are valid without "appl" prefix
• Ignoring prefix existence
• Choosing shortest word instead of longest

4. What is the bug in this Trie insertion snippet for finding the longest word?

for (let char of word) {
  if (!node[char]) {
    node[char] = {};
  }
  node = node[char];
}
node.isWord = true;

medium

A. No bug, code is correct

B. Missing initialization of root node

C. Not marking the end of word properly

D. Using plain object instead of TrieNode class

Solution

1. Step 1: Analyze insertion logic

   The loop creates child nodes if missing and moves down the Trie correctly.

2. Step 2: Check end of word marking

   Setting node.isWord = true marks the end of the word properly.

3. Final Answer:

   No bug, code is correct -> Option A

4. Quick Check:

   Insertion and marking done correctly [OK]

Hint: Mark end of word after insertion loop [OK]

Common Mistakes:

• Forgetting to mark end of word
• Overcomplicating node creation
• Assuming class needed when object works

5. You have a list of words: ["w", "wo", "wor", "worl", "world", "banana", "ban", "bana", "banan"]. Using a Trie, which word is the longest that can be built one character at a time from other words in the list?

hard

A. "banan"

B. "world"

C. "banana"

D. "worl"

Solution

1. Step 1: Verify prefix existence for each candidate

   For "world": prefixes "w", "wo", "wor", "worl", "world" all exist. For "banana": prefixes "b", "ba", "ban", "bana", "banan", "banana" -- but "b" and "ba" are missing. Similarly, "banan" misses "b" and "ba".

2. Step 2: Compare longest buildable words

   "world" is the longest word where all prefixes from length 1 to full length exist in the list.

3. Final Answer:

   "world" -> Option B

4. Quick Check:

   Longest buildable word = "world" [OK]

Hint: Check all prefixes exist before choosing longest word [OK]

Common Mistakes:

• Choosing "banana" without all prefixes
• Ignoring missing prefixes
• Confusing partial prefixes as complete