Word Search in Trie in DSA C - Time & Space Complexity

Time Complexity: Word Search in Trie

O(n)

Understanding Time Complexity

When searching for a word in a trie, we want to know how the time taken grows as the word length changes.

We ask: How many steps does it take to find a word in the trie?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Search for a word in a trie
bool search(TrieNode* root, const char* word) {
    TrieNode* current = root;
    for (int i = 0; word[i] != '\0'; i++) {
        int index = word[i] - 'a';
        if (current->children[index] == NULL) {
            return false;
        }
        current = current->children[index];
    }
    return current->isEndOfWord;
}
    

This code checks each letter of the word in the trie nodes one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop over each character of the input word.
• How many times: Exactly once per character, so as many times as the word length.

How Execution Grows With Input

Each additional letter in the word adds one more step to check in the trie.

Input Size (n)	Approx. Operations
10	10 checks
100	100 checks
1000	1000 checks

Pattern observation: The number of operations grows directly with the word length.

Final Time Complexity

Time Complexity: O(n)

This means the search time grows linearly with the length of the word being searched.

Common Mistake

[X] Wrong: "Searching a word in a trie takes time proportional to the number of words stored."

[OK] Correct: The search depends only on the length of the word, not how many words are in the trie.

Interview Connect

Understanding this linear time search helps you explain efficient prefix lookups and autocomplete features clearly.

Self-Check

"What if the trie stored uppercase and lowercase letters separately? How would the time complexity change?"

Practice

1. What is the main purpose of using a Trie data structure in word search?

easy

A. To efficiently store and search words by their prefixes

B. To sort words alphabetically

C. To store words in a hash table

D. To perform arithmetic operations on words

Solution

1. Step 1: Understand Trie structure

   A Trie stores characters of words in a tree-like structure where each node represents a character.

2. Step 2: Identify Trie usage

   It allows quick prefix-based search by traversing nodes matching characters in order.

3. Final Answer:

   To efficiently store and search words by their prefixes -> Option A

4. Quick Check:

   Trie = prefix search [OK]

Hint: Trie is for prefix-based word search, not sorting [OK]

Common Mistakes:

• Confusing Trie with hash tables
• Thinking Trie sorts words
• Assuming Trie stores whole words in one node

2. Which of the following is the correct way to check if a word exists in a Trie in C?

easy

A. Use a hash function on the word and check hash table

B. Check only the root node for the word

C. Sort the Trie nodes and do binary search

D. Traverse nodes for each character; if path ends early, word not found

Solution

1. Step 1: Understand Trie search logic

   Search each character by moving through child nodes; if any character node missing, word not found.

2. Step 2: Eliminate incorrect options

   Hash and sorting are unrelated to Trie search; root node alone can't hold whole word.

3. Final Answer:

   Traverse nodes for each character; if path ends early, word not found -> Option D

4. Quick Check:

   Trie search = character traversal [OK]

Hint: Check each character node in order to find word [OK]

Common Mistakes:

• Using hash or sorting instead of Trie traversal
• Checking only root node
• Assuming word is stored in one node

3. What will be the output of the following C code snippet that searches for the word "cat" in a Trie containing words "cat", "car", "dog"?

bool search(TrieNode* root, const char* word) {
    TrieNode* node = root;
    for (int i = 0; word[i] != '\0'; i++) {
        int index = word[i] - 'a';
        if (!node->children[index]) return false;
        node = node->children[index];
    }
    return node->isEndOfWord;
}

// Assume Trie is built with words "cat", "car", "dog"
printf("%d", search(root, "cat"));

medium

A. 1

B. 0

C. Compilation error

D. Runtime error

Solution

1. Step 1: Analyze search function logic

   Function traverses Trie nodes for each character; returns true if last node marks end of word.

2. Step 2: Check if "cat" is in Trie

   "cat" was inserted, so traversal succeeds and isEndOfWord is true, returning 1.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Word found = 1 [OK]

Hint: If word inserted, search returns 1 (true) [OK]

Common Mistakes:

• Assuming search returns 0 for found words
• Confusing isEndOfWord flag
• Ignoring null checks in traversal

4. Identify the bug in this Trie search function in C:

bool search(TrieNode* root, const char* word) {
    TrieNode* node = root;
    for (int i = 0; word[i] != '\0'; i++) {
        int index = word[i] - 'a';
        if (node->children[index] == NULL) return false;
        node = node->children[index];
    }
    return true;
}

medium

A. It uses wrong index calculation for characters

B. It does not check for NULL root pointer

C. It returns true without checking if node marks end of word

D. It should return false at the end instead of true

Solution

1. Step 1: Review return condition after loop

   Function returns true after traversal without verifying if current node marks end of a valid word.

2. Step 2: Understand Trie word end marking

   Words end at nodes with isEndOfWord true; missing this check causes false positives.

3. Final Answer:

   It returns true without checking if node marks end of word -> Option C

4. Quick Check:

   Must check isEndOfWord before returning true [OK]

Hint: Always check isEndOfWord after traversal [OK]

Common Mistakes:

• Skipping isEndOfWord check
• Not handling NULL root
• Incorrect index calculation

5. Given a Trie containing the words: "apple", "app", "ape", "bat", "bath", which of the following words will the search function return true for? Assume the search function correctly checks isEndOfWord.

hard

A. "ap" only

B. "bat", "bath", and "apple" only

C. "bat", "bathroom", and "apple" only

D. "bat" and "apple" only

Solution

1. Step 1: Identify words inserted in Trie

   Words inserted are "apple", "app", "ape", "bat", "bath"; only these have isEndOfWord true at last character.

2. Step 2: Check each option word

   "ap" is prefix but not a word; "bat", "bath", "apple" are present; "bathroom" not inserted.

3. Final Answer:

   "bat", "bath", and "apple" only -> Option B

4. Quick Check:

   Search true only for inserted words [OK]

Hint: Search true only for complete inserted words [OK]

Common Mistakes:

• Assuming prefixes are valid words
• Including words not inserted
• Ignoring isEndOfWord flag