Trie Search Operation in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node marks the end of a word.

DSA Rust

if node.is_end_of_word [1] { return true; }

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

A== true

B== false

C= false

D= true

Attempts:

3 left

💡 Hint

Common Mistakes

Using assignment '=' instead of comparison '=='

Checking for false instead of true

The search should return true if the node's is_end_of_word flag is true, indicating a complete word.

2fill in blank

medium

Complete the code to get the index of the child node for the current character.

DSA Rust

let index = (word.chars().nth(i).unwrap() as usize) [1] ('a' as usize);

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

A+

B/

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Adding instead of subtracting

Multiplying or dividing instead of subtracting

To find the child index for a character, subtract the ASCII value of 'a' from the character's ASCII value.

3fill in blank

hard

Fix the error in the code to check if the child node exists before continuing the search.

DSA Rust

if node.children[index] [1] None { return false; }

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

A!=

B==

C=

D!==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '=='

Using assignment '=' instead of comparison

We must check if the child node is None to know if the path exists. Using '==' compares correctly.

4fill in blank

hard

Fill both blanks to correctly move to the child node and continue the search.

DSA Rust

node = node.children[[1]].as_ref().unwrap();
i += [2];

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

Aindex

B1

C0

Dword.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or word.len() instead of 1 for increment

Using wrong variable for child index

We use the calculated index to access the child node and increment i by 1 to move to the next character.

5fill in blank

hard

Fill all three blanks to complete the search function that returns true if the word exists in the trie.

DSA Rust

fn search(&self, word: &str) -> bool {
    let mut node = &self.root;
    let mut i = 0;
    while i < word.len() {
        let index = (word.chars().nth(i).unwrap() as usize) [1] ('a' as usize);
        if node.children[index] [2] None {
            return false;
        }
        node = node.children[index].as_ref().unwrap();
        i [3];
    }
    node.is_end_of_word == true
}

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

A-

B==

C+= 1

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong operators for index calculation

Using '!=' instead of '==' for None check

Not incrementing i correctly

The search subtracts 'a' from the character to get the index, checks if the child is None using '==', and increments i by 1 to move forward.

Practice

1. What does the search operation in a Trie data structure do?

easy

A. Counts the total words stored in the Trie

B. Inserts a new word into the Trie

C. Checks if a given word exists in the Trie

D. Deletes a word from the Trie

Solution

1. Step 1: Understand the purpose of search in Trie

   The search operation is designed to check if a specific word is present in the Trie structure.

2. Step 2: Differentiate from other operations

   Insertion adds words, deletion removes words, and counting totals words, but search only checks existence.

3. Final Answer:

   Checks if a given word exists in the Trie -> Option C

4. Quick Check:

   Search = Check existence [OK]

Hint: Search checks presence, not insertion or deletion [OK]

Common Mistakes:

• Confusing search with insert operation
• Thinking search deletes words
• Assuming search counts words

2. Which of the following is the correct Rust function signature for a Trie search method that returns a boolean?

easy

A. fn search(&self, word: &str) -> bool

B. fn search(word: String) -> bool

C. fn search(&mut self, word: &str) -> String

D. fn search(&self, word: &str) -> usize

Solution

1. Step 1: Identify correct method receiver and parameters

   The search method should borrow self immutably (&self) and take a string slice (&str) for the word to search.

2. Step 2: Confirm return type

   Search returns a boolean indicating presence, so return type must be bool.

3. Final Answer:

   fn search(&self, word: &str) -> bool -> Option A

4. Quick Check:

   Immutable borrow + &str + bool return [OK]

Hint: Search uses &self and returns bool [OK]

Common Mistakes:

• Using mutable borrow &mut self unnecessarily
• Returning wrong type like String or usize
• Taking owned String instead of &str

3. Given the following Rust Trie search code snippet, what will be the output when searching for "cat"?

let mut trie = Trie::new();
trie.insert("car");
trie.insert("cat");
trie.insert("cap");
println!("{}", trie.search("cat"));

medium

A. Compilation error

B. false

C. panic at runtime

D. true

Solution

1. Step 1: Analyze inserted words

   The words "car", "cat", and "cap" are inserted into the Trie, so "cat" is present.

2. Step 2: Check search output for "cat"

   Searching "cat" should return true because it was inserted.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Inserted words include "cat" so search returns true [OK]

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

Common Mistakes:

• Assuming search returns false if similar words exist
• Confusing insert and search methods
• Expecting runtime errors without reason

4. Identify the bug in this Rust Trie search method snippet:

fn search(&self, word: &str) -> bool {
    let mut node = &self.root;
    for ch in word.chars() {
        if !node.children.contains_key(&ch) {
            return false;
        }
        node = node.children.get(&ch).unwrap();
    }
    true
}

medium

A. Uses unwrap() which causes panic always

B. Returns true even if word is a prefix but not complete word

C. Does not iterate over characters correctly

D. Incorrect return type

Solution

1. Step 1: Understand search logic

   The code checks if each character exists in children nodes, moving down the Trie.

2. Step 2: Check final return condition

   It returns true after loop without verifying if current node marks end of a complete word.

3. Step 3: Identify bug

   This causes search to return true for prefixes that are not full words.

4. Final Answer:

   Returns true even if word is a prefix but not complete word -> Option B

5. Quick Check:

   Missing end-of-word check causes false positives [OK]

Hint: Check end-of-word flag after traversal [OK]

Common Mistakes:

• Ignoring end-of-word marker in search
• Assuming unwrap() always panics here
• Misunderstanding iteration over chars

5. You want to modify a Trie search operation in Rust to support searching with wildcard character '.' which can match any single letter. Which approach is best to implement this feature?

hard

A. Use recursion to explore all children nodes when '.' is encountered

B. Replace '.' with all alphabets before searching iteratively

C. Ignore '.' and treat it as a normal character

D. Use a hash map to store wildcard positions and skip them

Solution

1. Step 1: Understand wildcard '.' meaning

   The '.' can match any single character, so search must consider all possible children at that position.

2. Step 2: Choose search strategy

   Recursion allows exploring all children nodes when '.' is found, checking all paths.

3. Step 3: Evaluate other options

   Replacing '.' with all alphabets is inefficient; ignoring '.' is incorrect; hash map for positions doesn't solve matching.

4. Final Answer:

   Use recursion to explore all children nodes when '.' is encountered -> Option A

5. Quick Check:

   Recursion handles wildcards by branching [OK]

Hint: Use recursion to handle '.' wildcard in search [OK]

Common Mistakes:

• Ignoring wildcard meaning
• Trying to replace '.' with all letters before search
• Skipping recursive exploration