Why Longest Word in Dictionary Using Trie in DSA Typescript?

The Big Idea

What if you could find the longest word built one letter at a time without checking every prefix manually?

The Scenario

Imagine you have a huge list of words and you want to find the longest word where every prefix of that word is also in the list. Doing this by checking each word and its prefixes one by one feels like searching for a needle in a haystack.

The Problem

Manually checking each word and all its prefixes is slow and tiring. You might miss some prefixes or repeat checks many times. It's like repeatedly flipping pages in a big book to find a pattern, which wastes time and causes mistakes.

The Solution

A Trie is like a smart tree that stores words by their letters. It lets you quickly check if prefixes exist without repeating work. Using a Trie, you can easily find the longest word where all prefixes are present by walking through the tree step by step.

Before vs After

✗ Before

for (const word of words) {
  let allPrefixesExist = true;
  for (let i = 1; i < word.length; i++) {
    if (!wordsSet.has(word.substring(0, i))) {
      allPrefixesExist = false;
      break;
    }
  }
  if (allPrefixesExist) updateLongestWord(word);
}

✓ After

class TrieNode {
  children: Map<string, TrieNode> = new Map();
  isWord: boolean = false;
}

// Insert words into Trie and find longest word by DFS
function findLongestWord(words: string[]): string {
  // Build Trie
  // DFS to find longest word where all prefixes are words
}

What It Enables

This method lets you quickly find the longest word with all prefixes present, even in huge word lists, saving time and effort.

Real Life Example

Think of a dictionary app that suggests the longest valid word you can build letter by letter, helping players in word games like Scrabble or crossword puzzles.

Key Takeaways

Manual prefix checks are slow and repetitive.

Trie stores words by letters for fast prefix lookup.

Using Trie, finding the longest valid word becomes efficient and reliable.

Practice

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

easy

A. To efficiently store and search words by their prefixes

B. To sort words alphabetically

C. To count the total number of words

D. To reverse the words quickly

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 A

4. Quick Check:

   Trie = prefix search [OK]

Hint: Trie is best for prefix-based word searches [OK]

Common Mistakes:

• Thinking Trie sorts words alphabetically
• Confusing Trie with simple arrays
• Assuming Trie counts words only

2. Which of the following TypeScript code snippets correctly initializes a Trie node with children and an end-of-word flag?

easy

A. class TrieNode { children: Map = new Map(); isEnd: boolean = false; }

B. class TrieNode { children: string[] = []; isEnd: number = 0; }

C. class TrieNode { children: object = {}; isEnd: string = 'false'; }

D. class TrieNode { children: Map = new Map(); isEnd: boolean = true; }

Solution

1. Step 1: Check children type

   Children should map characters (string) to TrieNode objects, so Map is correct.

2. Step 2: Check isEnd flag type

   isEnd should be boolean to mark word end, so boolean = false is correct.

3. Final Answer:

   class TrieNode { children: Map = new Map(); isEnd: boolean = false; } -> Option A

4. Quick Check:

   Correct types = class TrieNode { children: Map = new Map(); isEnd: boolean = false; } [OK]

Hint: Use Map and boolean for TrieNode [OK]

Common Mistakes:

• Using array instead of Map for children
• Using wrong types for isEnd flag
• Mixing key and value types in Map

3. Given the following TypeScript code snippet that inserts words into a Trie and finds the longest word, what will be the output?

const words = ["a", "app", "apple", "apply", "ap"];
const trie = new Trie();
words.forEach(word => trie.insert(word));
console.log(trie.findLongestWord());

medium

A. "a"

B. "apply"

C. "apple"

D. "app"

Solution

1. Step 1: Insert words into Trie

   Words "a", "app", "apple", "apply", "ap" are inserted. Trie stores prefixes for all.

2. Step 2: Find longest word with all prefixes present

   "app" has all prefixes present: "a", "ap", "app". "apple" and "apply" require "appl" which is not a word in the list.

3. Final Answer:

   "app" -> Option D

4. Quick Check:

   Longest word with all prefixes = "app" [OK]

Hint: Choose longest word with all prefixes present, lex smallest if tie [OK]

Common Mistakes:

• Choosing "apple" or "apply" without checking if "appl" is a word
• Ignoring that intermediate prefixes like "appl" must be full words
• Picking the overall longest word without prefix validation

4. Consider this buggy TypeScript code snippet for inserting words into a Trie:

insert(word: string) {
  let node = this.root;
  for (const char of word) {
    node = node.children.get(char);
    if (!node.children.has(char)) {
      node.children.set(char, new TrieNode());
    }
  }
  node.isEnd = true;
}

What is the main issue causing a runtime error?

medium

A. The loop does not iterate over all characters

B. The isEnd flag is set before inserting nodes

C. node.children.get(char) can return undefined, causing node to be undefined

D. The children map is not initialized

Solution

1. Step 1: Analyze node update inside loop

   node.children.get(char) may return undefined if key missing, but code assumes it returns TrieNode.

2. Step 2: Identify cause of runtime error

   If undefined assigned to node, next iteration fails accessing children of undefined.

3. Final Answer:

   node.children.get(char) can return undefined, causing node to be undefined -> Option C

4. Quick Check:

   Undefined node from get() causes error [OK]

Hint: Check if get() can return undefined before assignment [OK]

Common Mistakes:

• Ignoring possibility of undefined from Map.get()
• Assuming loop skips characters
• Confusing isEnd flag placement

5. You have a list of words: ["w", "wo", "wor", "worl", "world", "banana", "ban", "bana", "banan"]. Using a Trie, which word will be the longest word in the dictionary where every prefix is also a word in the list?

hard

A. "banana"

B. "world"

C. "banan"

D. "worl"

Solution

1. Step 1: Check prefixes for "world"

   Prefixes: "w", "wo", "wor", "worl" all exist in list.

2. Step 2: Check prefixes for "banana" and "banan"

   "banana" and "banan" are missing shorter prefixes like "b" and "ba" which are not in the list.

3. Step 3: Compare valid longest words

   "world" is longest with all prefixes present; others longer but invalid.

4. Final Answer:

   "world" -> Option B

5. Quick Check:

   Longest word with all prefixes = "world" [OK]

Hint: Check all prefixes exist; longest valid word wins [OK]

Common Mistakes:

• Choosing "banana" ignoring missing prefix like "b"
• Confusing similar prefixes
• Not verifying all prefixes exist