Trie Insert Operation in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add a character node to the children map.

DSA Typescript

if (!node.children.has(char)) {
  node.children.[1](char, new TrieNode());
}

Drag options to blanks, or click blank then click option'

Aset

Badd

Cpush

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using array methods like push instead of Map methods.

Trying to use add which is for Set, not Map.

The Map object uses the set method to add a key-value pair. Here, we add a new character node.

2fill in blank

medium

Complete the code to move to the next node for the current character.

DSA Typescript

node = node.children.[1](char)!;

Drag options to blanks, or click blank then click option'

Adelete

Bset

Chas

Dget

Attempts:

3 left

💡 Hint

Common Mistakes

Using set instead of get, which would overwrite the node.

Using has which returns a boolean, not the node.

The get method retrieves the value for a given key in a Map. Here, it gets the child node for the character.

3fill in blank

hard

Fix the error in marking the end of a word in the Trie node.

DSA Typescript

node.[1] = true;

Drag options to blanks, or click blank then click option'

AwordEnd

BisEnd

CendOfWord

DisWord

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent property names that do not exist in the TrieNode class.

Using properties that are not boolean flags.

The property isEnd is commonly used to mark the end of a word in a Trie node.

4fill in blank

hard

Fill both blanks to complete the loop that inserts each character of the word.

DSA Typescript

for (const [1] of word) {
  if (!node.children.[2]([1])) {
    node.children.set([1], new TrieNode());
  }
  node = node.children.get([1])!;
}

Drag options to blanks, or click blank then click option'

Achar

Bletter

Chas

Dincludes

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'includes' which is an array or string method, not Map.

Using 'letter' as loop variable but 'has' is the correct Map method.

The loop variable is commonly named char to represent each character. The has method checks if the character node exists in the children map.

5fill in blank

hard

Fill all three blanks to complete the insert method for the Trie.

DSA Typescript

insert(word: string): void {
  let node = this.root;
  for (const [1] of word) {
    if (!node.children.[2]([1])) {
      node.children.set([1], new TrieNode());
    }
    node = node.children.get([1])!;
  }
  node.[3] = true;
}

Drag options to blanks, or click blank then click option'

Achar

Bhas

CisEnd

Dletter

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect property names for the end of word flag.

Using string methods instead of Map methods for children.

The loop variable is char. The has method checks for the character in children. The isEnd property marks the end of the word.

Practice

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

easy

A. To add a word character by character into the Trie

B. To delete a word from the Trie

C. To search for a word in the Trie

D. To count the number of words in the Trie

Solution

1. Step 1: Understand the insert operation

   The insert operation adds each character of a word into the Trie one by one, creating new nodes if needed.

2. Step 2: Differentiate from other operations

   Deleting removes words, searching finds words, counting totals words. Insert specifically adds words.

3. Final Answer:

   To add a word character by character into the Trie -> Option A

4. Quick Check:

   Insert = Add word [OK]

Hint: Insert means adding word step-by-step [OK]

Common Mistakes:

• Confusing insert with search or delete
• Thinking insert counts words
• Assuming insert replaces words

2. Which of the following is the correct TypeScript syntax to start inserting a word into a Trie node named root?

easy

A. let node = root;

B. node = root;

C. const node = root()

D. var node == root;

Solution

1. Step 1: Understand variable declaration in TypeScript

   Using let declares a variable that can be reassigned, suitable for traversal nodes.

2. Step 2: Check syntax correctness

   let node = root; uses correct assignment with let. node = root; misses declaration keyword. const node = root() wrongly calls root as a function. var node == root; uses invalid comparison operator.

3. Final Answer:

   let node = root; -> Option A

4. Quick Check:

   Correct variable declaration = let node = root; [OK]

Hint: Use let for variable assignment in TypeScript [OK]

Common Mistakes:

• Missing let or const keyword
• Using == instead of = for assignment
• Calling node as a function

3. Given the following TypeScript code snippet for inserting the word "cat" into an empty Trie, what will be the printed output after insertion?

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

const root = new TrieNode();

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

insert("cat");

console.log(Array.from(root.children.keys()));

medium

A. ["cat"]

B. ["c", "a", "t"]

C. []

D. ["c"]

Solution

1. Step 1: Trace insertion of "cat" into Trie

   Starting at root, 'c' is not present, so a new node for 'c' is created. Then move to 'c' node, 'a' is added as child, then 't' as child of 'a'.

2. Step 2: Check root's children keys

   Root only has one child node for 'c'. The keys of root.children are ['c'].

3. Final Answer:

   ["c"] -> Option D

4. Quick Check:

   Root children keys after insert = ["c"] [OK]

Hint: Root children keys show first letters only [OK]

Common Mistakes:

• Expecting all letters at root level
• Confusing keys with full words
• Ignoring Map keys method

4. Identify the error in the following TypeScript code snippet for inserting a word into a Trie:

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

medium

A. Using 'let' instead of 'const' for node

B. Missing parentheses when creating new TrieNode

C. Not checking if node.children.get(char) is undefined

D. Not marking isEndOfWord as false initially

Solution

1. Step 1: Check new object creation syntax

   In TypeScript, when creating a new object, parentheses are required even if no arguments are passed. new TrieNode should be new TrieNode().

2. Step 2: Verify other parts

   Using 'let' for node is fine since node changes. The get method is safe here because we just set the key if missing. isEndOfWord is set at the end, initial value doesn't cause error.

3. Final Answer:

   Missing parentheses when creating new TrieNode -> Option B

4. Quick Check:

   new object needs () [OK]

Hint: Always use () when calling constructors [OK]

Common Mistakes:

• Omitting () after new keyword
• Confusing let and const for variables
• Ignoring possible undefined from get()

5. You want to insert multiple words ["bat", "ball", "batman"] into a Trie. After inserting all, which of the following statements is true about the Trie structure?

hard

A. The node for 'b' has children 'a' and 'm' directly

B. Each word creates a separate branch from root

C. The node for 'b' has children 'a' only, and 'a' has children 't' and 'l'

D. The node for 't' is the end of word for both "bat" and "batman"

Solution

1. Step 1: Analyze common prefixes in words

   Words "bat", "ball", and "batman" share prefix 'b' -> 'a'. From 'a', 't' leads to "bat" and "batman", 'l' leads to "ball".

2. Step 2: Understand Trie branching

   Node 'b' has child 'a'. Node 'a' branches to 't' and 'l'. 'm' is child of 'a'->'t'->'m' for "batman". So 'm' is not direct child of 'b'.

3. Final Answer:

   The node for 'b' has children 'a' only, and 'a' has children 't' and 'l' -> Option C

4. Quick Check:

   Trie branches at shared prefixes [OK]

Hint: Words share nodes for common prefixes only [OK]

Common Mistakes:

• Assuming all words start new branches at root
• Thinking 'm' is direct child of 'b'
• Confusing end of word nodes