Word Search in Trie in DSA Typescript

Mental Model

A trie stores words by sharing common letter paths, so searching a word means following its letters step-by-step down the trie branches.

Analogy: Imagine a tree of roads where each road sign is a letter. To find a word, you follow the signs letter by letter until you reach the end of the word or get lost.

root
 ↓
 [c] -> [a] -> [t*]
        ↓
       [r] -> [a] -> [t*]

* means end of a word

Dry Run Walkthrough

Input: Insert words: cat, car, rat; Search word: car

Goal: Find if the word 'car' exists in the trie

Step 1: Start at root, check for 'c'

root -> [c] -> ...

Why: We begin searching from the root for the first letter

Step 2: Move to node 'c', check for 'a'

root -> [c] -> [a] -> ...

Why: Next letter 'a' must be a child of 'c'

Step 3: Move to node 'a', check for 'r'

root -> [c] -> [a] -> [r*]

Why: Next letter 'r' must be a child of 'a'

Step 4: Node 'r' marks end of word, word found

root -> [c] -> [a] -> [r*]

Why: End marker confirms 'car' is stored

Result:

root -> [c] -> [a] -> [r*]
Word 'car' found: true

Annotated Code

DSA Typescript

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

class Trie {
  root: TrieNode = new TrieNode();

  insert(word: string): void {
    let node = this.root;
    for (const char of word) {
      if (!node.children.has(char)) {
        node.children.set(char, new TrieNode());
      }
      node = node.children.get(char)!; // advance to child node
    }
    node.isEndOfWord = true; // mark end of word
  }

  search(word: string): boolean {
    let node = this.root;
    for (const char of word) {
      if (!node.children.has(char)) {
        return false; // letter path missing
      }
      node = node.children.get(char)!; // advance to child node
    }
    return node.isEndOfWord; // check end marker
  }
}

// Driver code
const trie = new Trie();
trie.insert('cat');
trie.insert('car');
trie.insert('rat');
const wordToSearch = 'car';
console.log(`Word '${wordToSearch}' found:`, trie.search(wordToSearch));

node = node.children.get(char)!; // advance to child node

advance node pointer to next letter node

node.isEndOfWord = true; // mark end of word

mark current node as end of a stored word

if (!node.children.has(char)) { return false; }

stop search if next letter path missing

return node.isEndOfWord; // check end marker

confirm word ends exactly here

Output

Word 'car' found: true

Complexity Analysis

Time: O(m) because we check each of the m letters in the word once during search

Space: O(n*m) where n is number of words and m average length, for storing all nodes

vs Alternative: Compared to searching in a list of words (O(n*m)), trie search is faster as it avoids checking all words by branching on letters

Edge Cases

Search for empty string

Returns false since no word is empty

DSA Typescript

return node.isEndOfWord; // check end marker

Search for word not inserted

Returns false when letter path missing

DSA Typescript

if (!node.children.has(char)) { return false; }

Insert duplicate word

No error, marks end of word again

DSA Typescript

node.isEndOfWord = true; // mark end of word

When to Use This Pattern

When you need fast prefix or whole word search among many words, use a trie to follow letter paths efficiently.

Common Mistakes

Mistake: Not marking the end of word node, causing search to wrongly find prefixes as words
Fix: Set isEndOfWord = true at the last node of inserted word

Mistake: Returning true if all letters found but ignoring end marker
Fix: Check isEndOfWord before confirming word exists

Summary

Stores words by branching on letters to enable fast search.

Use when you want quick word or prefix lookup among many words.

The key is following letter nodes step-by-step and checking end markers.

Practice

1. What is the main purpose of using a Trie data structure in word search problems?

easy

A. To efficiently store and search words by their prefixes

B. To sort words alphabetically

C. To store words in a linked list format

D. To perform arithmetic operations on words

Solution

1. Step 1: Understand the Trie structure

   A Trie stores characters in a tree where each path represents a prefix of words.

2. Step 2: Identify the use in word search

   This structure allows quick prefix-based search, making it efficient for word search problems.

3. Final Answer:

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

4. Quick Check:

   Trie = prefix search [OK]

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

Common Mistakes:

• Confusing Trie with sorting structures
• Thinking Trie stores words linearly
• Assuming Trie performs arithmetic

2. Which of the following is the correct TypeScript syntax to declare a Trie node class with a children map and an end-of-word boolean?

easy

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

B. class TrieNode { children: string[]; isEndOfWord: number; constructor() { this.children = []; this.isEndOfWord = 0; } }

C. class TrieNode { children: Map; isEndOfWord: boolean; constructor() { this.children = new Map(); this.isEndOfWord = true; } }

D. class TrieNode { children: object; isEndOfWord: string; constructor() { this.children = {}; this.isEndOfWord = 'false'; } }

Solution

1. Step 1: Check children type

   Children should be a Map from string to TrieNode for efficient character mapping.

2. Step 2: Check isEndOfWord type and initialization

   It should be a boolean initialized to false to mark word endings correctly.

3. Final Answer:

   class TrieNode { children: Map; isEndOfWord: boolean; constructor() { this.children = new Map(); this.isEndOfWord = false; } } -> Option A

4. Quick Check:

   Correct types and initialization [OK]

Hint: Use Map and boolean for end flag [OK]

Common Mistakes:

• Using array instead of Map for children
• Using wrong types for isEndOfWord
• Initializing isEndOfWord incorrectly

3. Given the following TypeScript code snippet for searching a word in a Trie, what will be the output when searching for the word "cat" if only "car" and "cap" are inserted?

class TrieNode {
  children = new Map();
  isEndOfWord = false;
}

class Trie {
  root = new TrieNode();

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

  search(word: string): boolean {
    let node = this.root;
    for (const ch of word) {
      if (!node.children.has(ch)) return false;
      node = node.children.get(ch)!;
    }
    return node.isEndOfWord;
  }
}

const trie = new Trie();
trie.insert("car");
trie.insert("cap");
console.log(trie.search("cat"));

medium

A. true

B. false

C. undefined

D. Runtime error

Solution

1. Step 1: Trace insertion of "car" and "cap"

   Both words create nodes for 'c', 'a', then diverge at 'r' and 'p' respectively, marking those ends as true.

2. Step 2: Trace search for "cat"

   Search finds 'c' and 'a', but 't' is not a child of 'a', so search returns false.

3. Final Answer:

   false -> Option B

4. Quick Check:

   Word not inserted = false [OK]

Hint: If last char missing, search returns false [OK]

Common Mistakes:

• Assuming partial prefix means true
• Confusing isEndOfWord with node existence
• Expecting undefined instead of false

4. Identify the bug in the following TypeScript Trie search method that causes it to always return false:

search(word: string): boolean {
  let node = this.root;
  for (const ch of word) {
    if (!node.children.has(ch)) return false;
    node = node.children.get(ch)!;
  }
  return node.isEndOfWord === false;
}

medium

A. The method should return true before the loop ends

B. The loop should iterate backwards over the word

C. The children map should be an array instead

D. The return condition should be node.isEndOfWord, not its negation

Solution

1. Step 1: Analyze the return statement

   The method returns node.isEndOfWord === false, which means it returns true if the word is NOT ending here.

2. Step 2: Correct the return logic

   It should return node.isEndOfWord to confirm the word ends at this node.

3. Final Answer:

   The return condition should be node.isEndOfWord, not its negation -> Option D

4. Quick Check:

   Return true if word ends here [OK]

Hint: Return node.isEndOfWord directly to confirm word end [OK]

Common Mistakes:

• Returning negation of isEndOfWord
• Changing loop direction unnecessarily
• Using wrong data structure for children

5. You want to implement a function that finds all words in a Trie starting with a given prefix. Which approach correctly combines Trie traversal and backtracking to collect all matching words?

hard

A. Search each word in the Trie individually to check prefix match

B. Store all words in an array and filter by prefix each time

C. Traverse Trie nodes for prefix, then recursively collect words from that node using backtracking

D. Use a hash map to map prefixes to words without Trie traversal

Solution

1. Step 1: Traverse Trie to prefix node

   Follow characters of prefix to reach the node representing last prefix character.

2. Step 2: Use backtracking from prefix node

   Recursively explore all children nodes to collect complete words starting with prefix.

3. Final Answer:

   Traverse Trie nodes for prefix, then recursively collect words from that node using backtracking -> Option C

4. Quick Check:

   Prefix traversal + backtracking = Traverse Trie nodes for prefix, then recursively collect words from that node using backtracking [OK]

Hint: Find prefix node, then backtrack to collect words [OK]

Common Mistakes:

• Checking all words instead of prefix traversal
• Filtering array instead of Trie traversal
• Using hash map without Trie structure