Trie Search Operation in DSA C - Time & Space Complexity

Time Complexity: Trie Search Operation

O(n)

Understanding Time Complexity

We want to know how long it takes to find a word in a trie.

How does the search time change when the word gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Search for a word in a trie
bool trieSearch(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 node->isEndOfWord;
}
    

This code checks each letter of the word step-by-step in the trie.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop over each character of the input word.
• How many times: Once for each letter in the word (length n).

How Execution Grows With Input

As the word gets longer, the search checks more letters one by one.

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

Pattern observation: The number of steps 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.

Common Mistake

[X] Wrong: "Searching a word in a trie depends on the total number of words stored."

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

Interview Connect

Knowing trie search time helps you explain fast word lookups in real apps like autocomplete or spell check.

Self-Check

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

Practice

1. What does the search operation in a Trie data structure primarily check for?

easy

A. The number of words stored in the Trie

B. If the Trie is empty

C. If a given word exists in the Trie

D. The length of the longest word in the Trie

Solution

1. Step 1: Understand the purpose of Trie search

   The search operation in a Trie is designed to check if a specific word is present by traversing nodes corresponding to each character.

2. Step 2: Identify what the search confirms

   If all characters of the word are found sequentially and the last node marks the end of a word, the word exists in the Trie.

3. Final Answer:

   If a given word exists in the Trie -> Option C

4. Quick Check:

   Trie search checks word existence = C [OK]

Hint: Search checks presence of full word in Trie [OK]

Common Mistakes:

• Confusing search with counting words
• Thinking search checks Trie emptiness
• Assuming search returns word length

2. Which of the following is the correct way to check if a character node exists in a Trie during search in C?

easy

A. if (node->children[c] == NULL)

B. if (node->children[c - 'a'] != NULL)

C. if (node->children[c + 'a'] != NULL)

D. if (node->children[c] != 0)

Solution

1. Step 1: Understand character indexing in Trie

   Characters are mapped to array indices by subtracting 'a' to get 0-based index for children array.

2. Step 2: Check for non-null child node

   To verify if a child node exists for character c, check if node->children[c - 'a'] is not NULL.

3. Final Answer:

   if (node->children[c - 'a'] != NULL) -> Option B

4. Quick Check:

   Index = c - 'a', check != NULL = B [OK]

Hint: Use c - 'a' to index children array in Trie [OK]

Common Mistakes:

• Using c directly as index without subtracting 'a'
• Checking for NULL with wrong comparison
• Adding 'a' to c instead of subtracting

3. Given the following Trie search code snippet in C, what will be the output if we search for "cat" after inserting "car" and "cat"?

bool search(struct TrieNode* root, const char* word) {
    struct 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 node->isEndOfWord;
}

medium

A. true

B. false

C. Compilation error

D. Runtime error

Solution

1. Step 1: Understand inserted words and search target

   Words "car" and "cat" are inserted, so nodes for 'c', 'a', 'r' and 't' exist with isEndOfWord true at 'r' and 't'.

2. Step 2: Trace search for "cat"

   Search traverses 'c' -> 'a' -> 't' nodes, all present, and node->isEndOfWord is true at 't', so returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Inserted "cat" found = true [OK]

Hint: Search returns true if word was inserted before [OK]

Common Mistakes:

• Assuming search returns false if similar word exists
• Confusing isEndOfWord with node presence
• Ignoring null checks in loop

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

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

medium

A. Infinite loop due to missing increment

B. Does not check for NULL root

C. Incorrect indexing of children array

D. Returns false even if word exists

Solution

1. Step 1: Analyze return statement after loop

   The function returns false after loop ends, ignoring if current node marks end of word.

2. Step 2: Identify missing check for word completion

   It should return node->isEndOfWord to confirm full word presence, but returns false always.

3. Final Answer:

   Returns false even if word exists -> Option D

4. Quick Check:

   Missing isEndOfWord check causes false return [OK]

Hint: Return node->isEndOfWord after loop to confirm word [OK]

Common Mistakes:

• Forgetting to check isEndOfWord at end
• Assuming node presence means word exists
• Ignoring root NULL check (less critical here)

5. You have a Trie storing words: "bat", "bath", "baton", "bar", and "barn". Which of the following search queries will return false? "bathe", "bar", "bat", "barn"

hard

A. "bathe"

B. "bar"

C. "bat"

D. "barn"

Solution

1. Step 1: Check each query word against inserted words

   "bathe" is not inserted; "bar", "bat", and "barn" are inserted words in the Trie.

2. Step 2: Understand Trie search returns false for non-inserted words

   Since "bathe" is not inserted, search will fail at 'h' or after, returning false.

3. Final Answer:

   "bathe" -> Option A

4. Quick Check:

   Non-inserted word returns false = "bathe" [OK]

Hint: Search false if word not fully inserted in Trie [OK]

Common Mistakes:

• Assuming prefix words always return true
• Confusing prefix presence with full word presence
• Ignoring isEndOfWord flag in search