Why Trie Exists and What Hash Map Cannot Do for Strings in DSA Javascript - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a word into the Trie node children.

DSA Javascript

node.children[[1]] = new TrieNode();

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

Aword

Bchar

Cindex

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the whole word as a key instead of a single character.

Using the node itself as a key.

Each character (char) of the word is used as a key in the children map to create or access the next Trie node.

2fill in blank

medium

Complete the code to check if the current node has a child for the character.

DSA Javascript

if (!node.children.hasOwnProperty([1])) { return false; }

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

Aword

Bnode

Cindex

Dchar

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for the whole word instead of a character.

Using an incorrect variable like index.

We check if the children object has the current character as a key to continue traversal.

3fill in blank

hard

Fix the error in the search function to correctly return if a word exists in the Trie.

DSA Javascript

return node.isEndOfWord === [1];

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

Afalse

Bnull

Ctrue

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Returning false instead of true for end of word.

Returning null or undefined which are incorrect.

The search should return true only if the node marks the end of a word.

4fill in blank

hard

Fill both blanks to complete the prefix search function that returns true if any word starts with the prefix.

DSA Javascript

for (let [1] = 0; [1] < prefix.length; [1]++) {
  let [2] = prefix[[1]];
  if (!node.children.hasOwnProperty([2])) return false;
  node = node.children[[2]];
}
return true;

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

Ai

Bchar

Cindex

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same variable name for both index and character.

Using incorrect variable names that don't match the code.

The loop index variable is 'i' and the current character is 'char' used to check children.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each word to its length only if the length is greater than 3.

DSA Javascript

const result = { [1]: [2] for (const [3] of words) if [2] > 3 };

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

Aword

Bword.length

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect variable names or values in the comprehension.

Confusing keys and values.

We map each word (key) to its length (value) for words longer than 3 characters.