Trie Search Operation in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

if (!node.children.has([1])) {
    return false;
}

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Aindex

Bchar

Cword

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the whole word instead of a single character.

Using the node variable instead of the character.

We check if the current character 'char' exists in the children map of the node.

2fill in blank

medium

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

DSA Typescript

node = node.children.get([1])!;

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Achar

Bindex

Cword

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the whole word or index instead of the character.

Trying to get children with the node variable.

We get the child node corresponding to the current character 'char' to continue traversal.

3fill in blank

hard

Fix the error in the return statement to check if the node marks the end of a word.

DSA Typescript

return node.[1];

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AisWord

Bend

CisEndOfWord

DwordEnd

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect property names like 'isWord' or 'end'.

Returning the node itself instead of a boolean.

The property 'isEndOfWord' indicates if the node marks the end of a valid word in the trie.

4fill in blank

hard

Fill both blanks to iterate over each character in the word and check the trie.

DSA Typescript

for (let [1] = 0; [2] < word.length; [2]++) {
    const char = word[[1]];
    // rest of the code
}

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Ai

Bj

Cindex

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in initialization and condition.

Using variables not declared in the loop.

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

5fill in blank

hard

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

DSA Typescript

function search(word: string): boolean {
    let node = this.root;
    for (let i = 0; i < word.length; i++) {
        const char = word[[1]];
        if (!node.children.has([2])) {
            return false;
        }
        node = node.children.get([3])!;
    }
    return node.isEndOfWord;
}

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

Ai

Bchar

Cword

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'word' instead of 'char' to check children.

Using 'i' instead of 'char' to get children.

We use 'i' to get the character from 'word', then check and get the child node using 'char'.