Longest Word in Dictionary Using Trie in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a word into the Trie.

DSA Javascript

insert(word) {
  let node = this.root;
  for (let char of word) {
    if (!node.children.has(char)) {
      node.children.set(char, new TrieNode());
    }
    node = node.[1];
  }
  node.isEnd = true;
}

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

Achildren.has(char)

Bchildren.set(char)

Cchildren.get(char)

Dchildren.delete(char)

Attempts:

3 left

💡 Hint

Common Mistakes

Using set instead of get to move to the child node.

Checking if child exists but not moving to it.

We move to the child node corresponding to the current character using children.get(char).

2fill in blank

medium

Complete the code to check if a word exists in the Trie.

DSA Javascript

search(word) {
  let node = this.root;
  for (let char of word) {
    if (!node.children.has(char)) {
      return false;
    }
    node = node.children.get(char);
  }
  return node[1];
}

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

A.isWord

B.found

C.exists

D.isEnd

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong property name like isWord or exists.

Returning true without checking if it's the end of a word.

The isEnd property tells if the node marks the end of a valid word.

3fill in blank

hard

Fix the error in the DFS function to find the longest word built one character at a time.

DSA Javascript

dfs(node, path) {
  if (!node.isEnd && node !== this.root) return;
  if (path.length > this.longest.length) {
    this.longest = path;
  }
  for (let [char, child] of node.children) {
    this.dfs(child, path + [1]);
  }
}

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

Achild

Bchar

Cnode

Dpath

Attempts:

3 left

💡 Hint

Common Mistakes

Adding the child node object instead of the character.

Adding the whole node or path again.

We add the character char to the current path string during DFS.

4fill in blank

hard

Fill both blanks to complete the reduce function that builds an object filtering words with length greater than 3 and mapping them to their lengths.

DSA Javascript

const result = words.reduce((acc, word) => {
  if (word.length [2] 3) {
    acc[word][1] word.length;
  }
  return acc;
}, {});

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

A:

B>

C<

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using colon instead of equals for object property assignment.

Using less than sign for filtering.

The equals sign = assigns the length as the value for the word key in the accumulator object, and > filters words longer than 3.

5fill in blank

hard

Fill all three blanks to complete the function that finds the longest word in the dictionary using Trie and DFS.

DSA Javascript

function longestWord(words) {
  const trie = new Trie();
  for (const word of words) {
    trie.[1](word);
  }
  trie.longest = "";
  trie.[2](trie.root, "");
  return trie.[3];
}

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

Ainsert

Bdfs

Clongest

Dsearch

Attempts:

3 left

💡 Hint

Common Mistakes

Calling search instead of insert to add words.

Returning dfs instead of the longest word.

Insert words into the Trie, run DFS to find the longest word, then return the longest word.