Trie Search Operation in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current character exists in the children of the trie node.

DSA C

if (node->children[[1]] == NULL) {
    return 0;
}

Drag options to blanks, or click blank then click option'

Aword[i] - 'a'

Bword[i]

Ci

D'a'

Attempts:

3 left

💡 Hint

Common Mistakes

Using the character directly as index causes out-of-bound errors.

Using 'i' as index instead of character index.

We subtract 'a' from the character to get the index in the children array (0-25).

2fill in blank

medium

Complete the code to move to the next node in the trie for the current character.

DSA C

node = node->children[[1]];

Drag options to blanks, or click blank then click option'

Aword[i]

Bword[i] - 'a'

Ci

D'a'

Attempts:

3 left

💡 Hint

Common Mistakes

Using the character directly instead of index.

Using loop index 'i' instead of character index.

We move to the child node corresponding to the current character index.

3fill in blank

hard

Fix the error in the return statement to correctly check if the word ends here.

DSA C

return node->[1];

Drag options to blanks, or click blank then click option'

AisWordEnd

BendOfWord

CwordEnd

DisEndOfWord

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect field names causes compilation errors.

Confusing the flag name with similar variants.

The correct field name to check if a node marks the end of a word is 'isEndOfWord'.

4fill in blank

hard

Fill both blanks to complete the loop that iterates over each character of the word.

DSA C

for (int [1] = 0; [2] < length; [1]++) {
    // loop body
}

Drag options to blanks, or click blank then click option'

Ai

Bj

Clength

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables in loop condition and increment.

Using variable names that do not match the loop initialization.

The loop variable 'i' is used consistently to iterate from 0 to length-1.

5fill in blank

hard

Fill all three blanks to complete the search function that returns 1 if word is found, else 0.

DSA C

int search(struct TrieNode* root, const char* word) {
    struct TrieNode* node = root;
    int length = strlen(word);
    for (int [1] = 0; [1] < [2]; [1]++) {
        if (node->children[word[[1]] - 'a'] == NULL) {
            return 0;
        }
        node = node->children[word[[1]] - 'a'];
    }
    return node->isEndOfWord;
}

Drag options to blanks, or click blank then click option'

Ai

Blength

Cword

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong loop variable or limit causes incorrect iteration.

Accessing characters with wrong index variable.

The loop uses 'i' to iterate from 0 to length-1, and accesses characters with word[i].

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