Autocomplete System with Trie in DSA C++ - Time & Space Complexity

Time Complexity: Autocomplete System with Trie

O(k)

Understanding Time Complexity

We want to understand how fast an autocomplete system using a trie can find suggestions based on a prefix.

How does the time to find suggestions grow as the prefix or dictionary size changes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct TrieNode {
    TrieNode* children[26] = {nullptr};
    bool isWord = false;
};

void insert(TrieNode* root, const std::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->isWord = true;
}

TrieNode* searchPrefix(TrieNode* root, const std::string& prefix) {
    TrieNode* node = root;
    for (char c : prefix) {
        int idx = c - 'a';
        if (!node->children[idx])
            return nullptr;
        node = node->children[idx];
    }
    return node;
}
    

This code inserts words into a trie and searches for the node matching a prefix.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character of the input word or prefix.
• How many times: Once per character, so the number of characters in the word or prefix (let's call it k).

How Execution Grows With Input

As the prefix length k grows, the search checks one more character each time.

Input Size (prefix length k)	Approx. Operations
10	About 10 character checks
100	About 100 character checks
1000	About 1000 character checks

Pattern observation: The time grows linearly with the prefix length, not with the total number of words stored.

Final Time Complexity

Time Complexity: O(k)

This means the time to find the prefix node grows only with the length of the prefix, regardless of how many words are stored.

Common Mistake

[X] Wrong: "The search time depends on the total number of words stored in the trie."

[OK] Correct: The search only follows the prefix characters, so it depends on prefix length, not total words.

Interview Connect

Understanding this helps you explain why tries are efficient for autocomplete, showing you know how data structure choice affects speed.

Self-Check

"What if we also want to list all words starting with the prefix? How would that affect the time complexity?"

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