Why Trie Exists and What Hash Map Cannot Do for Strings in DSA C++ - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a character into the Trie node's children.

DSA C++

node->children[[1]] = new TrieNode();

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

Aword[i]

Bi

Cnode

Dchildren

Attempts:

3 left

💡 Hint

Common Mistakes

Using the index 'i' instead of the character.

Using the node pointer instead of a character key.

We use the current character of the word (word[i]) as the key to insert into the children map.

2fill in blank

medium

Complete the code to check if a prefix exists in the Trie.

DSA C++

if (node->children.find([1]) == node->children.end()) return false;

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

Anode

Bprefix[i]

Ci

Dchildren

Attempts:

3 left

💡 Hint

Common Mistakes

Using the index 'i' instead of the character.

Using the node pointer or children map incorrectly.

We check if the current character of the prefix exists in the children map.

3fill in blank

hard

Fix the error in the code to mark the end of a word in the Trie.

DSA C++

node->[1] = true;

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Aend

Bchildren

CisWord

DwordEnd

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'children' which is a map, not a boolean.

Using 'end' or 'wordEnd' which may not be defined.

The boolean flag 'isWord' marks the end of a valid word in the Trie.

4fill in blank

hard

Fill both blanks to complete the Trie search function for a word.

DSA C++

for (int i = 0; i < word.size(); i++) {
    if (node->children.find([1]) == node->children.end()) return false;
    node = node->children[[2]];
}

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

Aword[i]

Bi

Cnode

Dchildren

Attempts:

3 left

💡 Hint

Common Mistakes

Using different keys for find and indexing.

Using the index 'i' instead of the character.

Both blanks require the current character of the word to traverse the Trie correctly.

5fill in blank

hard

Fill all three blanks to complete the code that builds an unordered_map of words longer than 3 characters to their lengths.

DSA C++

std::unordered_map<std::string, int> lengths;
for (const auto& [3] : words) {
    if ([2] > 3) {
        lengths[[1]] = [2];
    }
}

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

Aword

Bword.length()

Cwords

Dw

Attempts:

3 left

💡 Hint

Common Mistakes

Using the collection name instead of the loop variable.

Using the wrong property for length.

We map each word to its length if the length is greater than 3. 'word' is the loop variable.