Word Search in Trie in DSA Typescript - Time & Space Complexity

Time Complexity: Word Search in Trie

O(n)

Understanding Time Complexity

We want to understand how long it takes to find a word in a Trie data structure.

The question is: how does the search time grow as the word length changes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

function searchWord(root: TrieNode, word: string): boolean {
  let current = root;
  for (const char of word) {
    if (!current.children.has(char)) return false;
    current = current.children.get(char)!;
  }
  return current.isEndOfWord;
}
    

This code checks if a given word exists in the Trie by following each character step-by-step.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character of the word.
• How many times: Exactly once per character in the word.

How Execution Grows With Input

Each additional character in the word adds one more step to check.

Input Size (n)	Approx. Operations
10	10 checks
100	100 checks
1000	1000 checks

Pattern observation: The number of steps grows directly with the word length.

Final Time Complexity

Time Complexity: O(n)

This means the search time grows linearly with the length of the word you are searching for.

Common Mistake

[X] Wrong: "Searching a word in a Trie takes constant time because it's a tree."

[OK] Correct: The search depends on the word length, so longer words take more steps, not always constant time.

Interview Connect

Understanding this helps you explain why Tries are efficient for prefix searches and how their search time depends on word length, a common topic in interviews.

Self-Check

"What if the Trie stored words with only lowercase letters, and we used an array instead of a Map for children? How would the time complexity change?"

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