Trie Insert Operation in DSA Go - Build from Scratch

Trie Insert Operation

📖 Scenario: You are building a simple word dictionary to quickly check if words exist. You will create a Trie data structure to store words efficiently.

🎯 Goal: Build a Trie and insert words into it step-by-step. Finally, print the Trie's root children keys to see the first letters stored.

📋 What You'll Learn

Create a TrieNode struct with a map for children and a boolean for end of word

Create a Trie struct with a root TrieNode

Add a method Insert(word string) to add words to the Trie

Insert given words and print the root children keys

💡 Why This Matters

🌍 Real World

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

💼 Career

Understanding Trie insertions is important for roles involving search engines, text processing, and efficient data retrieval.

1

Create TrieNode and Trie structs

Create a struct called TrieNode with a field children of type map[rune]*TrieNode and a field isEnd of type bool. Then create a struct called Trie with a field root of type *TrieNode. Finally, create a function NewTrieNode() that returns a pointer to a new TrieNode with children initialized as an empty map and isEnd set to false.

DSA Go

# Your code here

Hint

Define the structs and initialize the children map in NewTrieNode.

2

Initialize Trie with root node

Create a function NewTrie() that returns a pointer to a new Trie with its root initialized by calling NewTrieNode().

DSA Go

package main

import "fmt"

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

type Trie struct {
    root *TrieNode
}

func NewTrieNode() *TrieNode {
    return &TrieNode{children: make(map[rune]*TrieNode), isEnd: false}
}

// Your code here

func main() {
    // Step 1 code only
}

Hint

Create NewTrie to initialize the root node of the Trie.

3

Add Insert method to Trie

Add a method Insert(word string) to the Trie struct. This method should start from the root node and for each character in word, check if it exists in the current node's children. If not, create a new TrieNode and add it to children. Move to the child node and after the last character, mark isEnd as true.

DSA Go

package main

import "fmt"

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

type Trie struct {
    root *TrieNode
}

func NewTrieNode() *TrieNode {
    return &TrieNode{children: make(map[rune]*TrieNode), isEnd: false}
}

func NewTrie() *Trie {
    return &Trie{root: NewTrieNode()}
}

// Add Insert method to Trie
// Your code here

func main() {
    // Step 2 code only
}

Hint

Loop over each character, create nodes if missing, and mark the end of the word.

4

Insert words and print root children keys

In main(), create a new Trie by calling NewTrie(). Insert the words "go", "golang", and "git" using the Insert method. Then print the keys of the root node's children map as a slice of runes.

DSA Go

package main

import "fmt"

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

type Trie struct {
    root *TrieNode
}

func NewTrieNode() *TrieNode {
    return &TrieNode{children: make(map[rune]*TrieNode), isEnd: false}
}

func NewTrie() *Trie {
    return &Trie{root: NewTrieNode()}
}

func (t *Trie) Insert(word string) {
    current := t.root
    for _, ch := range word {
        if _, ok := current.children[ch]; !ok {
            current.children[ch] = NewTrieNode()
        }
        current = current.children[ch]
    }
    current.isEnd = true
}

func main() {
    // Insert words and print root children keys
    // Your code here
}

Hint

Insert the words and print the root children keys. The output should be a slice with the rune for 'g' (103).

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