Trie Insert Operation in DSA Go - Time & Space Complexity

Time Complexity: Trie Insert Operation

O(n)

Understanding Time Complexity

We want to understand how the time to insert a word into a trie changes as the word gets longer.

How does the number of steps grow when we add longer words?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func (t *TrieNode) Insert(word string) {
    node := t
    for _, ch := range word {
        if node.children[ch] == nil {
            node.children[ch] = &TrieNode{children: make(map[rune]*TrieNode)}
        }
        node = node.children[ch]
    }
    node.isEnd = true
}
    

This code inserts each character of a word into the trie, creating new nodes if needed.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character of the input word.
• How many times: Exactly once for each character in the word (length n).

How Execution Grows With Input

Each character in the word requires one step to check or create a node.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The steps grow directly with the word length, so doubling the word doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert grows linearly with the length of the word.

Common Mistake

[X] Wrong: "Inserting a word takes constant time because we just add nodes once."

[OK] Correct: Each character must be processed, so longer words take more time, not the same time.

Interview Connect

Understanding this helps you explain how tries efficiently handle many words, especially when words share prefixes.

Self-Check

"What if we changed the data structure for children from a map to a fixed-size array? How would the time complexity change?"

Practice

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

easy

A. To delete a word from the Trie

B. To add a new word by creating nodes for each character

C. To search for a word in the Trie

D. To count the total words stored in the Trie

Solution

1. Step 1: Understand the Insert operation

   The Insert operation adds a word by creating nodes for each character if they don't exist.

2. Step 2: Differentiate from other operations

   Deleting removes words, searching finds words, counting totals words. Insert specifically adds words.

3. Final Answer:

   To add a new word by creating nodes for each character -> Option B

4. Quick Check:

   Insert = Add word nodes [OK]

Hint: Insert means adding nodes for each character [OK]

Common Mistakes:

• Confusing insert with search or delete
• Thinking insert counts words
• Assuming insert modifies existing words

2. Which of the following is the correct way to declare a Trie node struct in Go for insertion?

easy

A. type TrieNode struct { children map[rune]*TrieNode; isEnd bool }

B. type TrieNode struct { children []int; isEnd int }

C. type TrieNode struct { children map[string]int; isEnd string }

D. type TrieNode struct { children []string; isEnd bool }

Solution

1. Step 1: Identify correct children type for TrieNode

   Children should map characters (rune) to TrieNode pointers for branching.

2. Step 2: Check isEnd type

   isEnd is a boolean indicating if node ends a word.

3. Final Answer:

   type TrieNode struct { children map[rune]*TrieNode; isEnd bool } -> Option A

4. Quick Check:

   Children map[rune]*TrieNode and isEnd bool [OK]

Hint: Children map with rune keys and bool isEnd [OK]

Common Mistakes:

• Using slice instead of map for children
• Wrong data types for children or isEnd
• Using string keys instead of rune

3. What will be the output after inserting the words "cat" and "car" into an empty Trie and then checking if the node for 'a' under 'c' has children?

medium

A. nil

B. false

C. true

D. runtime error

Solution

1. Step 1: Insert "cat" and "car" into Trie

   Both words share prefix 'c' -> 'a'. After insertion, node 'a' has children 't' and 'r'.

2. Step 2: Check if node 'a' under 'c' has children

   Since 'a' node has two children nodes ('t' and 'r'), it has children, so true.

3. Final Answer:

   true -> Option C

4. Quick Check:

   Shared prefix node has children = true [OK]

Hint: Shared prefix nodes have children after insert [OK]

Common Mistakes:

• Assuming node 'a' has no children
• Confusing node existence with children presence
• Expecting nil or error instead of true

4. Given this Go code snippet for inserting a word into a Trie, what is the bug?

func (t *TrieNode) Insert(word string) {
    node := t
    for _, ch := range word {
        if node.children[ch] == nil {
            node.children[ch] = &TrieNode{}
        }
        node = node.children[ch]
    }
    node.isEnd = true
}

medium

A. children map is not initialized before use

B. Loop variable ch should be int, not rune

C. isEnd should be set before loop

D. Function should return a value

Solution

1. Step 1: Check initialization of children map

   The children map must be initialized (e.g., make(map[rune]*TrieNode)) before inserting nodes.

2. Step 2: Identify impact of missing initialization

   Without initialization, accessing children[ch] causes runtime panic (nil map assignment).

3. Final Answer:

   children map is not initialized before use -> Option A

4. Quick Check:

   Uninitialized map causes runtime error [OK]

Hint: Always initialize map before inserting keys [OK]

Common Mistakes:

• Ignoring map initialization
• Misunderstanding rune vs int for keys
• Thinking isEnd needs early assignment

5. You want to insert the words ["go", "gone", "guild"] into a Trie. After insertion, how many children does the root node have, and why?

hard

A. 0, because root has no children

B. 3, because there are three words

C. 2, because 'go' and 'guild' share 'g' but 'gone' differs

D. 1, because all words start with 'g'

Solution

1. Step 1: Analyze first characters of all words

   All words start with the letter 'g', so root node will have one child for 'g'.

2. Step 2: Understand Trie root children count

   Root children count equals distinct first letters of inserted words, here only 'g'.

3. Final Answer:

   1, because all words start with 'g' -> Option D

4. Quick Check:

   Root children = distinct first letters = 1 [OK]

Hint: Root children = unique first letters of words [OK]

Common Mistakes:

• Counting words instead of unique first letters
• Assuming root has no children
• Confusing shared prefixes with root children count