Trie Search Operation in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current character exists in the children nodes.

DSA Javascript

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

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

Achar

Bnode

Cindex

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the whole word instead of the current character.

Using the node object instead of the character key.

The search checks if the current character char exists in the children of the current node.

2fill in blank

medium

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

DSA Javascript

node = node.children[[1]];

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Aword

Bchar

Cindex

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the whole word or index instead of the character.

Trying to assign node to itself.

We update node to the child node corresponding to the current character char.

3fill in blank

hard

Fix the error in the return statement to correctly check if the word ends here.

DSA Javascript

return node.[1] === true;

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AisEndOfWord

Bend

CisWord

DwordEnd

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect property names like isWord or end.

Checking the property incorrectly without === true.

The property isEndOfWord marks if the node represents the end of a valid word in the Trie.

4fill in blank

hard

Fill both blanks to complete the loop that iterates over each character in the word.

DSA Javascript

for (let [1] = 0; [2] < word.length; [1]++) {
  const char = word[[1]];
  // rest of the code
}

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Ai

Bj

Ck

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables in the loop condition and increment.

Using variable names that are not consistent.

The loop uses i as the index variable to iterate over the word characters.

5fill in blank

hard

Fill all three blanks to complete the search function that returns true if the word exists in the Trie.

DSA Javascript

function search(word) {
  let node = this.root;
  for (let i = 0; i < word.length; i++) {
    const [1] = word[i];
    if (!node.children.hasOwnProperty([2])) {
      return false;
    }
    node = node.children[[3]];
  }
  return node.isEndOfWord === true;
}

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

Achar

Bword

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for the character in different places.

Checking children with the whole word instead of the character.

The variable char holds the current character. We check if char exists in children and then move to that child node.

Practice

1. What does the search operation in a Trie data structure primarily check for?

easy

A. The length of the longest word in the Trie

B. If the Trie is empty

C. If a given word exists in the Trie

D. The number of words stored in the Trie

Solution

1. Step 1: Understand the purpose of Trie search

   The search operation in a Trie is designed to check if a specific word is present in the Trie by traversing nodes corresponding to each character.

2. Step 2: Identify what the search confirms

   It confirms the presence or absence of the entire word, not the count or length of words stored.

3. Final Answer:

   If a given word exists in the Trie -> Option C

4. Quick Check:

   Trie search checks word existence [OK]

Hint: Search checks full word presence in Trie nodes [OK]

Common Mistakes:

• Confusing search with counting words
• Thinking search returns word length
• Assuming search checks if Trie is empty

2. Which of the following is the correct way to check if a character node exists in a Trie during search in JavaScript?

easy

A. if (node.char !== null)

B. if (node[char] !== undefined)

C. if (node.has(char))

D. if (node.char === true)

Solution

1. Step 1: Understand Trie node structure in JavaScript

   Trie nodes are usually objects where keys are characters. Checking if a character exists uses property access and undefined check.

2. Step 2: Identify correct syntax for existence check

   Using node[char] !== undefined correctly checks if the character key exists in the node object.

3. Final Answer:

   if (node[char] !== undefined) -> Option B

4. Quick Check:

   Check node property with !== undefined [OK]

Hint: Use node[char] !== undefined to check character node [OK]

Common Mistakes:

• Using dot notation with variable char
• Using incorrect method like has() on object
• Checking for null instead of undefined

3. What will be the output of the following JavaScript code snippet?

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

class Trie {
  constructor() {
    this.root = new TrieNode();
  }

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

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

const trie = new Trie();
trie.insert('cat');
trie.insert('car');
console.log(trie.search('cat'));
console.log(trie.search('cap'));

medium

A. true false

B. false true

C. true true

D. false false

Solution

1. Step 1: Insert words 'cat' and 'car' into Trie

   Both words are inserted by creating nodes for each character and marking the last node's isEnd as true.

2. Step 2: Search for 'cat' and 'cap'

   'cat' exists fully with isEnd true, so search returns true. 'cap' is missing 'p' node, so search returns false.

3. Final Answer:

   true false -> Option A

4. Quick Check:

   Search 'cat' = true, 'cap' = false [OK]

Hint: Search returns true only if full word and isEnd true [OK]

Common Mistakes:

• Assuming partial matches return true
• Not setting isEnd flag on insert
• Confusing children object access

4. Identify the bug in the following Trie search method and select the correct fix:

search(word) {
  let node = this.root;
  for (const char of word) {
    if (node.children[char] === undefined) return false;
    node = node.children[char];
  }
  return true;
}

medium

A. Use 'for (let i=0; i

B. Change '===' to '!==' in the if condition

C. Initialize node with a new TrieNode inside the method

D. Return node.isEnd instead of true at the end

Solution

1. Step 1: Analyze the search method's return value

   The method returns true after traversing all characters without checking if the last node marks the end of a word.

2. Step 2: Correct the return to check word completion

   Return node.isEnd to confirm the word ends here, not just prefix exists.

3. Final Answer:

   Return node.isEnd instead of true at the end -> Option D

4. Quick Check:

   Check isEnd flag to confirm full word [OK]

Hint: Always check isEnd flag after traversal [OK]

Common Mistakes:

• Returning true without isEnd check
• Changing comparison operator incorrectly
• Unnecessary loop syntax changes

5. Given a Trie containing the words ['apple', 'app', 'bat', 'bath'], which of the following search queries will return true? Assume the search method correctly checks for full word presence. Select the correct combination.

hard

A. search('app'), search('bat'), search('bath')

B. search('appl'), search('bat'), search('bath')

C. search('apple'), search('bat'), search('batman')

D. search('app'), search('bath'), search('batman')

Solution

1. Step 1: Identify words present in the Trie

   The Trie contains 'apple', 'app', 'bat', and 'bath'. Only these exact words return true on search.

2. Step 2: Check each option's search queries

   search('app'), search('bat'), search('bath') searches 'app', 'bat', 'bath' all present. Others include partial or missing words like 'appl' or 'batman' which return false.

3. Final Answer:

   search('app'), search('bat'), search('bath') -> Option A

4. Quick Check:

   Only full words in Trie return true [OK]

Hint: Search true only for full words inserted in Trie [OK]

Common Mistakes:

• Assuming partial words return true
• Confusing prefixes with full words
• Including words not inserted in Trie