Trie Node Design and Initialization in DSA Javascript - Time & Space Complexity

Time Complexity: Trie Node Design and Initialization

O(1)

Understanding Time Complexity

We want to understand how much work it takes to create a single node in a trie.

This helps us see how the trie grows as we add words.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code creates a trie node with an empty children map and a flag for word end.

Identify Repeating Operations

Look for any loops or repeated steps inside the constructor.

• Primary operation: Creating an empty object and setting a boolean.
• How many times: This happens once per node creation.

How Execution Grows With Input

Each node creation takes the same small amount of work, no matter what.

Input Size (n)	Approx. Operations
1 node	2 simple steps
10 nodes	20 simple steps
100 nodes	200 simple steps

Pattern observation: The work grows directly with the number of nodes created.

Final Time Complexity

Time Complexity: O(1)

This means creating one trie node always takes the same small amount of time, no matter what.

Common Mistake

[X] Wrong: "Creating a trie node takes longer if the trie is big already."

[OK] Correct: Each node is made fresh with no loops, so its creation time does not depend on trie size.

Interview Connect

Knowing that node creation is quick helps you focus on where the real work happens when adding or searching words.

Self-Check

"What if the children were stored in an array of fixed size instead of an object? How would the time complexity change?"

Practice

1. What is the main purpose of a Trie node in a Trie data structure?

easy

A. To store the entire word in one node

B. To store integer values for sorting

C. To hold only the last character of a word

D. To store characters and links to child nodes representing prefixes

Solution

1. Step 1: Understand Trie node role

   A Trie node stores a character and links to child nodes that represent possible next characters in words.

2. Step 2: Compare options

   Only To store characters and links to child nodes representing prefixes correctly describes storing characters and child links for prefixes.

3. Final Answer:

   To store characters and links to child nodes representing prefixes -> Option D

4. Quick Check:

   Trie node = character + child links [OK]

Hint: Trie nodes hold characters and children, not full words [OK]

Common Mistakes:

• Thinking nodes store whole words
• Confusing Trie nodes with array elements
• Assuming nodes only store last characters

2. Which of the following is the correct way to initialize a Trie node in JavaScript with an empty children object and a boolean flag isEndOfWord set to false?

easy

A. const node = { children: null, isEndOfWord: true };

B. const node = { children: [], isEndOfWord: false };

C. const node = { children: {}, isEndOfWord: false };

D. const node = { children: {}, isEndOfWord: true };

Solution

1. Step 1: Identify correct children type

   Children should be an object to map characters to nodes, so {} is correct.

2. Step 2: Check isEndOfWord initial value

   Initially, isEndOfWord should be false because the node does not end a word yet.

3. Final Answer:

   const node = { children: {}, isEndOfWord: false }; -> Option C

4. Quick Check:

   Empty object + false flag = correct initialization [OK]

Hint: Children is an object, isEndOfWord starts false [OK]

Common Mistakes:

• Using array instead of object for children
• Setting isEndOfWord true at initialization
• Setting children to null

3. Consider this JavaScript code snippet for a Trie node:

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

const node = new TrieNode();
console.log(node.isEndOfWord);
console.log(Object.keys(node.children).length);

What will be the output?

medium

A. false
0

B. true
0

C. false
undefined

D. undefined
0

Solution

1. Step 1: Check isEndOfWord value after construction

   Constructor sets isEndOfWord to false, so console.log prints false.

2. Step 2: Check children keys count

   Children is an empty object, so Object.keys(node.children).length is 0.

3. Final Answer:

   false 0 -> Option A

4. Quick Check:

   isEndOfWord = false, children keys = 0 [OK]

Hint: Empty children means zero keys; isEndOfWord defaults false [OK]

Common Mistakes:

• Assuming isEndOfWord is true by default
• Expecting children keys length to be undefined
• Confusing object keys count with children length

4. The following code attempts to create a Trie node but has an error:

function TrieNode() {
  this.children = [];
  this.isEndOfWord = false;
}

What is the main problem with this code?

medium

A. children should be an object, not an array

B. isEndOfWord should be true initially

C. Function should return a value

D. Missing semicolon after isEndOfWord assignment

Solution

1. Step 1: Analyze children data type

   Children must be an object to map characters to nodes, not an array.

2. Step 2: Check other parts

   isEndOfWord false is correct; function constructors do not need explicit return; semicolons are optional in JS.

3. Final Answer:

   children should be an object, not an array -> Option A

4. Quick Check:

   Children = object, not array [OK]

Hint: Children must be object for character keys, not array [OK]

Common Mistakes:

• Thinking arrays can map characters
• Expecting function constructors to return explicitly
• Focusing on semicolon errors

5. You want to design a Trie node class in JavaScript that supports Unicode characters as keys in children and initializes isEndOfWord to false. Which of the following implementations correctly achieves this and allows adding child nodes dynamically?

hard

A. class TrieNode {
constructor() {
this.children = {};
this.isEndOfWord = true;
}
addChild(char) {
this.children[char] = new TrieNode();
}
}

B. class TrieNode {
constructor() {
this.children = new Map();
this.isEndOfWord = false;
}
addChild(char) {
if (!this.children.has(char)) {
this.children.set(char, new TrieNode());
}
}
}

C. class TrieNode {
constructor() {
this.children = [];
this.isEndOfWord = false;
}
addChild(char) {
this.children.push(char);
}
}

D. class TrieNode {
constructor() {
this.children = new Set();
this.isEndOfWord = false;
}
addChild(char) {
this.children.add(char);
}
}

Solution

1. Step 1: Check children data structure for Unicode support

   Using Map allows any string keys including Unicode, better than plain objects.

2. Step 2: Verify isEndOfWord initialization and addChild method

   isEndOfWord is false initially, and addChild adds new TrieNode only if missing, correctly updating children Map.

3. Step 3: Compare other options

   class TrieNode { constructor() { this.children = {}; this.isEndOfWord = true; } addChild(char) { this.children[char] = new TrieNode(); } } sets isEndOfWord true initially (wrong). class TrieNode { constructor() { this.children = []; this.isEndOfWord = false; } addChild(char) { this.children.push(char); } } uses array which can't map chars to nodes. class TrieNode { constructor() { this.children = new Set(); this.isEndOfWord = false; } addChild(char) { this.children.add(char); } } uses Set which stores chars but not nodes.

4. Final Answer:

   Implementation using Map with isEndOfWord false and addChild method -> Option B

5. Quick Check:

   Map + false flag + dynamic addChild = correct Trie node [OK]

Hint: Use Map for Unicode keys and false isEndOfWord [OK]

Common Mistakes:

• Using array or Set instead of Map or object
• Setting isEndOfWord true at start
• Not checking if child exists before adding