Trie Insert Operation in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node when the character is not found.

DSA Javascript

if (!node.children.hasOwnProperty(char)) {
  node.children[char] = [1];
}

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

A''

B[]

Cnull

D{}

Attempts:

3 left

💡 Hint

Common Mistakes

Using an array [] instead of an object {}

Assigning null or empty string instead of an object

We create a new empty object {} to represent the child node for the character.

2fill in blank

medium

Complete the code to move to the next node after inserting or finding the character.

DSA Javascript

node = node.children[[1]];

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

Achar

Bword

Cindex

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the whole word instead of the character

Using the index instead of the character

We use the current character 'char' to move to the child node.

3fill in blank

hard

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

DSA Javascript

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 like 'wordEnd' or 'endOfWord'

Forgetting to mark the end of the word

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 into the Trie.

DSA Javascript

for (let [1] = 0; [1] < word.[2]; [1]++) {
  const char = word[[1]];
  // insertion logic
}

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

Ai

Blength

Cj

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'j' instead of 'i' for the loop variable

Using 'size' instead of 'length' for string length

We use 'i' as the loop variable and 'length' to get the word's length.

5fill in blank

hard

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

DSA Javascript

insert(word) {
  let node = this.root;
  for (let [1] = 0; [1] < word.[2]; [1]++) {
    const char = word[[1]];
    if (!node.children.hasOwnProperty(char)) {
      node.children[char] = [3];
    }
    node = node.children[char];
  }
  node.isEnd = true;
}

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

Ai

Blength

C{}

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'null' instead of '{}'

Using wrong loop variable or property for length

We use 'i' as the loop index, 'length' for word length, and '{}' to create a new child node.

Practice

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

easy

A. To count the total words stored in the Trie

B. To delete a word from the Trie

C. To search for a word in the Trie

D. To add a new word character by character into 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 nodes if they don't exist.

2. Step 2: Compare with other operations

   Deleting removes words, searching finds words, and counting totals words, but insert specifically adds words.

3. Final Answer:

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

4. Quick Check:

   Insert = Add word step-by-step [OK]

Hint: Insert means adding word letters stepwise [OK]

Common Mistakes:

• Confusing insert with search or delete
• Thinking insert adds whole word as one node
• Assuming insert counts words

2. Which of the following is the correct way to start inserting a word "cat" into an empty Trie in JavaScript?

easy

A. for (let char of word) { if (!node.children[char]) node.children[char] = new TrieNode(); node = node.children[char]; }

B. for (let i = 0; i < word.length; i++) { node = node.children[word[i]]; }

C. while (word) { node.children[word[0]] = new TrieNode(); word = word.slice(1); }

D. node.children[word] = new TrieNode();

Solution

1. Step 1: Analyze the loop for insertion

   for (let char of word) { if (!node.children[char]) node.children[char] = new TrieNode(); node = node.children[char]; } loops over each character, checks if child node exists, creates if not, then moves to that child node.

2. Step 2: Check other options for correctness

   for (let i = 0; i < word.length; i++) { node = node.children[word[i]]; } does not create nodes if missing. while (word) { node.children[word[0]] = new TrieNode(); word = word.slice(1); } modifies the word string but does not move node properly. node.children[word] = new TrieNode(); tries to add whole word as one child.

3. Final Answer:

   for (let char of word) { if (!node.children[char]) node.children[char] = new TrieNode(); node = node.children[char]; } -> Option A

4. Quick Check:

   Loop chars, create missing nodes, move node [OK]

Hint: Loop chars, create missing nodes, move node [OK]

Common Mistakes:

• Not creating nodes when missing
• Assigning whole word as one child
• Not updating current node pointer

3. What will be the structure of the Trie after inserting the words ["bat", "ball"] in order? Consider each node stores children in an object and a boolean isEnd for word end.

medium

A. {"b":{"a":{"t":{"isEnd":false},"l":{"l":{"isEnd":true}}}}}

B. {"b":{"a":{"t":{"isEnd":true},"l":{"isEnd":true}}}}

C. {"b":{"a":{"t":{"isEnd":true},"l":{"l":{"isEnd":true}}}}}

D. {"b":{"a":{"t":{"isEnd":true}},"l":{"l":{"isEnd":true}}}}

Solution

1. Step 1: Insert "bat" into empty Trie

   Creates nodes b -> a -> t with t's isEnd true.

2. Step 2: Insert "ball" next

   Nodes b -> a exist, then add l -> l with last l's isEnd true.

3. Final Answer:

   {"b":{"a":{"t":{"isEnd":true},"l":{"l":{"isEnd":true}}}}} -> Option C

4. Quick Check:

   Both words share prefix 'ba', separate branches for 't' and 'll' [OK]

Hint: Shared prefix nodes, separate branches for different suffixes [OK]

Common Mistakes:

• Marking wrong nodes as word end
• Not creating second 'l' node for 'ball'
• Overwriting nodes incorrectly

4. Given this insert function snippet, what is the bug?

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

medium

A. Missing check for empty word input

B. node is assigned to node.children instead of node.children[char]

C. isEnd should be set before the loop

D. Children should be an array, not an object

Solution

1. Step 1: Check node update inside loop

   Inside loop, node is set to node.children (object), not node.children[char] (child node), so traversal breaks.

2. Step 2: Confirm correct pointer update

   Correct is node = node.children[char] to move down the Trie properly.

3. Final Answer:

   node is assigned to node.children instead of node.children[char] -> Option B

4. Quick Check:

   Update node to child node, not children object [OK]

Hint: Update node to child node, not children object [OK]

Common Mistakes:

• Assigning node to children object instead of child node
• Setting isEnd too early
• Ignoring empty input check

5. You want to insert the words ["apple", "app", "ape"] into a Trie and then check if the word "app" is a complete word in the Trie. Which approach correctly handles this scenario?

hard

A. Insert each word character by character, mark isEnd true at last character; then search for "app" checking isEnd at last node

B. Insert all words, but mark isEnd true only for the longest word; search ignores isEnd

C. Insert words as full strings in one node; search by matching full string keys

D. Insert words without marking isEnd; search by checking if path exists

Solution

1. Step 1: Insert words with isEnd marking

   Each word is inserted letter by letter; isEnd is set true at last letter node to mark complete words.

2. Step 2: Search for "app" with isEnd check

   Search traverses nodes for 'a'->'p'->'p' and checks isEnd true to confirm "app" is a complete word, not just prefix.

3. Final Answer:

   Insert each word character by character, mark isEnd true at last character; then search for "app" checking isEnd at last node -> Option A

4. Quick Check:

   Mark word ends, check isEnd to confirm full word [OK]

Hint: Mark word ends with isEnd, check isEnd when searching [OK]

Common Mistakes:

• Not marking isEnd for shorter words
• Treating prefixes as complete words without isEnd
• Storing whole words as single nodes