Trie Insert Operation in DSA Javascript - Build from Scratch

Trie Insert Operation

📖 Scenario: Imagine you are building a simple search helper that stores words so you can quickly check if a word exists later. You will create a Trie, a tree-like structure where each node represents a letter. This helps in fast word lookup.

🎯 Goal: You will build the insert operation for a Trie. This means adding words letter by letter into the Trie structure.

📋 What You'll Learn

Create a TrieNode class with a children object and an isEndOfWord boolean

Create a Trie class with a root node

Add an insert method to the Trie class that adds a word letter by letter

Print the root node's children keys after inserting words to show the Trie structure

💡 Why This Matters

🌍 Real World

Tries are used in autocomplete features, spell checkers, and IP routing to quickly find words or prefixes.

💼 Career

Understanding Tries helps in roles involving search engines, text processing, and efficient data retrieval.

1

Create TrieNode and Trie classes

Create a class called TrieNode with a constructor that initializes children as an empty object and isEndOfWord as false. Then create a class called Trie with a constructor that initializes root as a new TrieNode.

DSA Javascript

// Create TrieNode and Trie classes with constructors
// Your code here

Hint

Think of TrieNode as a box holding letters and a flag for word end. Trie starts with an empty root node.

2

Add insert method to Trie

Inside the Trie class, add a method called insert that takes a parameter word. Initialize a variable current to this.root.

DSA Javascript

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

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

    // Add insert method here
    // Your code here
}

Hint

Start the insert method by pointing to the root node. You will move through the Trie from here.

3

Complete insert method to add letters

In the insert method, use a for loop with variable char to go through each letter of word. For each char, if current.children[char] does not exist, set it to a new TrieNode(). Then update current to current.children[char]. After the loop, set current.isEndOfWord to true.

DSA Javascript

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

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

    insert(word) {
        let current = this.root;
        // Add letters of word to Trie
        // Your code here
    }
}

Hint

Check if the letter exists in children. If not, create a new node. Move current to that node. Mark the end of the word after the loop.

4

Insert words and print Trie root children keys

Create a new Trie instance called trie. Insert the words "cat", "car", and "dog" using the insert method. Then print Object.keys(trie.root.children) to show the first letters stored in the Trie.

DSA Javascript

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

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

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

// Create Trie instance, insert words, and print root children keys
// Your code here

Hint

Insert each word using insert. Then print the keys of trie.root.children to see the first letters stored.

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