Autocomplete System with Trie in DSA C++ - Execution Trace

Concept Flow - Autocomplete System with Trie

Start: Insert words into Trie

↓

For each word: For each char

↓

If char node missing: Create node

↓

Move to next char node

↓

Mark end of word

↓

End Insert

↓

Start Autocomplete Query

↓

Traverse Trie for prefix chars

↓

If prefix node missing: Return empty

↓

From prefix node: DFS to find all words

↓

Return list of words

Insert words character by character into the Trie, creating nodes as needed. For autocomplete, traverse the Trie by prefix, then collect all words below that prefix.

Execution Sample

DSA C++

Insert("cat")
Insert("car")
Insert("dog")
Autocomplete("ca")

Insert words 'cat', 'car', 'dog' into Trie, then find all words starting with 'ca'.

Execution Table

Step	Operation	Nodes in Trie	Pointer Changes	Visual State
1	Insert 'cat' char 'c'	Root -> c	Create node 'c' from root	root -> c -> null
2	Insert 'cat' char 'a'	Root -> c -> a	Create node 'a' from 'c'	root -> c -> a -> null
3	Insert 'cat' char 't'	Root -> c -> a -> t	Create node 't' from 'a'	root -> c -> a -> t*
4	Insert 'car' char 'c'	Root -> c -> a -> t*	Node 'c' exists, move to 'c'	root -> c -> a -> t*
5	Insert 'car' char 'a'	Root -> c -> a -> t*	Node 'a' exists, move to 'a'	root -> c -> a -> t*
6	Insert 'car' char 'r'	Root -> c -> a -> t*, r	Create node 'r' from 'a'	root -> c -> a -> t*, r*
7	Insert 'dog' char 'd'	Root -> c -> a -> t*, r*; d	Create node 'd' from root	root -> c -> a -> t*, r*; d -> null
8	Insert 'dog' char 'o'	Root -> c -> a -> t*, r*; d -> o	Create node 'o' from 'd'	root -> c -> a -> t*, r*; d -> o -> null
9	Insert 'dog' char 'g'	Root -> c -> a -> t*, r*; d -> o -> g	Create node 'g' from 'o'	root -> c -> a -> t*, r*; d -> o -> g*
10	Autocomplete 'ca' prefix traversal	Root -> c -> a -> t*, r*; d -> o -> g*	Traverse root->c->a	At node 'a' with children 't*', 'r*'
11	DFS from 'a' to find words	Root -> c -> a -> t*, r*; d -> o -> g*	Visit 't*' and 'r*' nodes	Found words: 'cat', 'car'
12	Return autocomplete results	Root -> c -> a -> t*, r*; d -> o -> g*	Return ['cat', 'car']	Autocomplete output: ['cat', 'car']

💡 Autocomplete ends after collecting all words under prefix node 'a'.

Variable Tracker

Variable	Start	After Step 3	After Step 6	After Step 9	After Step 12
Trie Nodes	root only	root -> c -> a -> t*	root -> c -> a -> t*, r*	root -> c -> a -> t*, r*; d -> o -> g*	No change (search complete)
Current Node	root	at 't' after inserting 'cat'	at 'r' after inserting 'car'	at 'g' after inserting 'dog'	at 'a' after prefix traversal
Autocomplete Results	empty	empty	empty	empty	['cat', 'car']

Key Moments - 3 Insights

Why do we create new nodes only if the character node doesn't exist?

Because existing nodes represent shared prefixes. See steps 4 and 5 where 'c' and 'a' nodes already exist, so we reuse them instead of creating duplicates.

How do we know when a word ends in the Trie?

We mark the last character node of a word with a special flag (shown as '*'). For example, 't*' and 'r*' in steps 3 and 6 indicate word ends.

What happens if the prefix is not found during autocomplete?

The traversal stops early and returns empty. In the execution table, step 10 shows traversal reaching node 'a'. If 'a' was missing, autocomplete would return empty immediately.

Visual Quiz - 3 Questions

Look at the execution table at step 6. Which new node is created?

A Node 'd' from root

B Node 'r' from 'a'

C Node 't' from 'a'

D Node 'o' from 'd'

💡 Hint

Check the 'Pointer Changes' and 'Visual State' columns at step 6.

At which step does the Trie first contain the complete word 'dog'?

A Step 7

B Step 8

C Step 9

D Step 10

💡 Hint

Look for the step where node 'g*' is created, marking the end of 'dog'.

If we autocomplete prefix 'do', what would be the output?

A ['dog']

B ['cat', 'car']

C []

D ['do']

💡 Hint

Refer to the Trie structure after step 9 and the DFS process in step 11.

Concept Snapshot

Trie stores words by characters in nodes.
Insert: create nodes for missing chars, mark word end.
Autocomplete: traverse prefix, then DFS to collect words.
Shared prefixes reuse nodes, saving space.
Word end marked by special flag on node.
Efficient prefix search for autocomplete.

Full Transcript

We build a Trie by inserting words character by character. Each character is a node linked from the previous. If a character node exists, we reuse it to share prefixes. The end of a word is marked specially. For autocomplete, we traverse the Trie nodes matching the prefix characters. If the prefix exists, we do a depth-first search from that node to find all complete words below it. This way, autocomplete returns all words starting with the prefix efficiently.

Practice

1. What is the main purpose of using a Trie in an autocomplete system?

easy

A. To efficiently store and search prefixes of words

B. To sort words alphabetically

C. To store words in a linked list

D. To perform arithmetic operations on words

Solution

1. Step 1: Understand the role of Trie in prefix search

   A Trie is a tree-like data structure that stores characters of words in a way that common prefixes share nodes.

2. Step 2: Connect Trie usage to autocomplete

   Autocomplete requires quick lookup of words starting with a given prefix, which Trie supports efficiently by traversing prefix nodes.

3. Final Answer:

   To efficiently store and search prefixes of words -> Option A

4. Quick Check:

   Trie = prefix search [OK]

Hint: Trie is all about prefix matching fast [OK]

Common Mistakes:

• Confusing Trie with sorting structures
• Thinking Trie stores words linearly
• Assuming Trie does arithmetic

2. Which of the following is the correct way to declare a Trie node class in C++?

easy

A. class TrieNode { public: map children; bool isEnd; };

B. class TrieNode { int children[26]; bool isEnd; };

C. class TrieNode { public: TrieNode* children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = nullptr; isEnd = false; } };

D. struct TrieNode { vector<int> children; bool isEnd; };

Solution

1. Step 1: Check for pointer array initialization

   class TrieNode { public: TrieNode* children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = nullptr; isEnd = false; } }; correctly declares an array of 26 pointers to TrieNode, initialized to nullptr, representing children for each alphabet letter.

2. Step 2: Verify isEnd boolean initialization

   class TrieNode { public: TrieNode* children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = nullptr; isEnd = false; } }; sets isEnd to false in constructor, which marks if a word ends at this node.

3. Final Answer:

   class TrieNode { public: TrieNode* children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = nullptr; isEnd = false; } }; -> Option C

4. Quick Check:

   Pointer array + bool flag = class TrieNode { public: TrieNode* children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = nullptr; isEnd = false; } }; [OK]

Hint: Trie nodes use pointer arrays for children [OK]

Common Mistakes:

• Using int array instead of pointer array
• Not initializing children pointers
• Using map<int> instead of map<char>

3. What will be the output of the following C++ code snippet that inserts words into a Trie and searches for prefix "app"?

class TrieNode {
public:
    TrieNode* children[26];
    bool isEnd;
    TrieNode() {
        for(int i=0; i<26; i++) children[i] = nullptr;
        isEnd = false;
    }
};

class Trie {
public:
    TrieNode* root;
    Trie() { root = new TrieNode(); }

    void insert(string word) {
        TrieNode* node = root;
        for(char c : word) {
            int idx = c - 'a';
            if(!node->children[idx])
                node->children[idx] = new TrieNode();
            node = node->children[idx];
        }
        node->isEnd = true;
    }

    bool startsWith(string prefix) {
        TrieNode* node = root;
        for(char c : prefix) {
            int idx = c - 'a';
            if(!node->children[idx]) return false;
            node = node->children[idx];
        }
        return true;
    }
};

int main() {
    Trie trie;
    trie.insert("apple");
    trie.insert("app");
    cout << (trie.startsWith("app") ? "Yes" : "No") << endl;
    cout << (trie.startsWith("apl") ? "Yes" : "No") << endl;
    return 0;
}

medium

A. No
Yes

B. Yes
No

C. Yes
Yes

D. No
No

Solution

1. Step 1: Insert words "apple" and "app" into Trie

   Both words share prefix "app", so nodes for 'a', 'p', 'p' exist. "apple" adds nodes for 'l' and 'e'.

2. Step 2: Check startsWith for "app" and "apl"

   "app" prefix exists, so returns true (prints "Yes"). "apl" prefix does not exist (no 'l' after 'p'), so returns false (prints "No").

3. Final Answer:

   Yes No -> Option B

4. Quick Check:

   Prefix "app" found, "apl" not found = Yes No [OK]

Hint: Check prefix nodes existence for startsWith [OK]

Common Mistakes:

• Confusing isEnd with prefix existence
• Assuming "apl" is found because 'a' and 'p' exist
• Printing boolean instead of Yes/No

4. Identify the bug in this Trie insertion code snippet:

void insert(string word) {
    TrieNode* node = root;
    for(char c : word) {
        int idx = c - 'a';
        if(node->children[idx] == nullptr)
            node->children[idx] = new TrieNode();
        node = node->children[idx];
        node->isEnd = true;
    }
}

medium

A. Setting isEnd = true inside the loop causes all prefixes to be marked as words

B. Not initializing children array to nullptr

C. Missing root initialization

D. Incorrect index calculation for children array

Solution

1. Step 1: Analyze isEnd assignment inside loop

   Setting isEnd = true inside the loop marks every node along the word as an end of a word, incorrectly marking prefixes as complete words.

2. Step 2: Correct placement of isEnd assignment

   isEnd should be set to true only after the loop ends, marking only the last node as end of the inserted word.

3. Final Answer:

   Setting isEnd = true inside the loop causes all prefixes to be marked as words -> Option A

4. Quick Check:

   isEnd inside loop = wrong word marking [OK]

Hint: Set isEnd true only after full word insertion [OK]

Common Mistakes:

• Marking prefixes as complete words
• Forgetting to set isEnd at all
• Incorrect children pointer checks

5. You want to implement an autocomplete system that returns all words starting with a given prefix using a Trie. Which approach correctly retrieves all such words?

hard

A. Use a hash map to store prefix to word list mapping directly

B. Store all words in a vector and filter by prefix each time

C. Traverse Trie and return words only if isEnd is true at prefix node

D. Traverse Trie to prefix node, then perform DFS to collect all words below it

Solution

1. Step 1: Locate prefix node in Trie

   Traverse the Trie character by character to reach the node representing the last character of the prefix.

2. Step 2: Perform DFS from prefix node to collect words

   From this node, recursively explore all child nodes to find all words that start with the prefix, collecting those where isEnd is true.

3. Final Answer:

   Traverse Trie to prefix node, then perform DFS to collect all words below it -> Option D

4. Quick Check:

   Prefix node + DFS = all autocomplete words [OK]

Hint: Find prefix node, then DFS to get all words [OK]

Common Mistakes:

• Returning only prefix node word ignoring children
• Filtering all words each query (inefficient)
• Using hash map without Trie traversal