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 map.

DSA C++

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

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

Aword

B0

Cnode

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using the whole word instead of an index.

Using a variable not defined in the loop.

We use 'i' as the index to access each character in the word during traversal.

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'

Ai

Bnode

Cword[i]

Dchildren

Attempts:

3 left

💡 Hint

Common Mistakes

Using the index 'i' directly instead of the character.

Using 'node' or 'children' which are not keys.

We use word[i] to get the current character and move to its child node.

3fill in blank

hard

Fix the error in the return statement to check if the word ends at this node.

DSA C++

return node->[1];

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

AisEnd

BisWord

CendOfWord

DwordEnd

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect flag names like 'isWord' or 'endOfWord'.

Forgetting to check this flag.

The boolean flag 'isEnd' indicates if the node marks the end of a word.

4fill in blank

hard

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

DSA C++

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

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

Ai

Bj

Ck

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in the loop condition and increment.

Using a variable not declared.

The loop variable 'i' is used consistently to iterate over the word characters.

5fill in blank

hard

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

DSA C++

bool search(string word) {
    TrieNode* node = root;
    for (int [1] = 0; [2] < word.size(); ++[1]) {
        if (node->children.find(word[[3]]) == node->children.end()) {
            return false;
        }
        node = node->children[word[[3]]];
    }
    return node->isEnd;
}

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

Ai

Bj

Ck

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for loop and indexing.

Using a constant index instead of the loop variable.

The loop variable 'i' is used consistently to index characters in the word during traversal.