Trie Insert Operation in DSA Javascript - Execution Trace

Concept Flow - Trie Insert Operation

Start at root node

↓

For each character in word

↓

Check if child node exists

↓

Move to child

↓

Mark end of word at last char

↓

Done

Insert each character by moving through existing nodes or creating new ones, then mark the last node as end of word.

Execution Sample

DSA Javascript

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

// Insert 'cat' into trie

This code defines the TrieNode class used in the insert operation for the word 'cat', as simulated in the execution table.

Execution Table

Step	Operation	Current Character	Node Created	Pointer Changes	Trie Visual State
1	Start at root	-	No	current = root	root {children: {}, isEndOfWord: false}
2	Check 'c' child	c	Yes	root.children['c'] = new TrieNode()	root {c} → node_c {children: {}, isEndOfWord: false}
3	Move to 'c' node	c	No	current = node_c	root {c} → node_c
4	Check 'a' child	a	Yes	node_c.children['a'] = new TrieNode()	root {c} → node_c {a} → node_a
5	Move to 'a' node	a	No	current = node_a	root {c} → node_c {a} → node_a
6	Check 't' child	t	Yes	node_a.children['t'] = new TrieNode()	root {c} → node_c {a} → node_a {t} → node_t
7	Move to 't' node	t	No	current = node_t	root {c} → node_c {a} → node_a {t} → node_t
8	Mark end of word	-	No	node_t.isEndOfWord = true	root {c} → node_c {a} → node_a {t} → node_t*
9	Done	-	No	-	Trie contains 'cat'

💡 All characters processed and last node marked as end of word

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	After Step 8	Final
current	root	node_c	node_a	node_t	node_t	node_t
node_c.children	{}	{}	{a: node_a}	{a: node_a}	{a: node_a}	{a: node_a}
node_a.children	{}	{}	{}	{t: node_t}	{t: node_t}	{t: node_t}
node_t.isEndOfWord	false	false	false	false	true	true

Key Moments - 3 Insights

Why do we create a new node only if the child does not exist?

Because if the child node for a character already exists (see steps 3, 5, 7), we just move to it to avoid overwriting existing words.

Why do we mark isEndOfWord only after processing the last character?

Marking isEndOfWord at the last node (step 8) tells the trie that a complete word ends here, distinguishing prefixes from full words.

What happens if we insert a word that shares prefix with existing words?

We reuse existing nodes for shared characters and only create new nodes for new characters, preserving the trie structure efficiently.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of 'current' after step 4?

A node_a

B node_c

C root

D node_t

💡 Hint

Check the 'variable_tracker' row for 'current' after step 4

At which step is the isEndOfWord flag set to true?

A Step 6

B Step 8

C Step 4

D Step 2

💡 Hint

Look at the 'execution_table' row with 'Mark end of word' operation

If the word 'car' was inserted after 'cat', which step would create a new node?

A For character 'c'

B For character 'a'

C For character 'r'

D No new nodes needed

💡 Hint

Shared prefix nodes exist for 'c' and 'a', new node needed only for differing character

Concept Snapshot

Trie Insert Operation:
- Start at root node
- For each character:
  - If child exists, move to it
  - Else create new child node
- After last char, mark node as end of word
- Efficiently stores words sharing prefixes

Full Transcript

The Trie Insert Operation starts at the root node. For each character in the word, it checks if a child node exists. If yes, it moves to that node; if no, it creates a new node. After processing all characters, it marks the last node as the end of a word. This process allows efficient storage of words with shared prefixes by reusing existing nodes and only adding new nodes when necessary.

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