Prefix matching with tries in Data Structures Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a word into a trie node.

Data Structures Theory

node.children[[1]] = TrieNode()

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Achar

Bindex

Cword

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the whole word instead of a single character as the key.

Using an index number instead of the character.

The variable char represents the current character in the word being inserted into the trie. We use it as the key to add a new child node.

2fill in blank

medium

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

Data Structures Theory

if [1] not in node.children:
    return False

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Aprefix

Bword

Cchar

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for the whole prefix string instead of one character.

Using the node variable instead of the character.

We check if the current character char of the prefix exists among the children of the current node.

3fill in blank

hard

Fix the error in the code to move to the next node during prefix search.

Data Structures Theory

node = node.children[[1]]

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Achar

Bprefix

Cword

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using the entire prefix string as a key.

Using an index number instead of the character.

We update node to the child node corresponding to the current character char in the prefix.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length if the word starts with 'pre'.

Data Structures Theory

{word: [1] for word in words if word.[2]('pre')}

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Alen(word)

Bstartswith

C==

Din

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of 'startswith' to check prefix.

Using 'in' which checks substring anywhere, not just prefix.

The dictionary comprehension maps each word to its length using len(word). The condition uses word.startswith('pre') to filter words starting with 'pre'.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their lengths if the length is greater than 3.

Data Structures Theory

{ [1]: [2] for [3] in words if len(word) > 3 }

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Aword.upper()

Blen(word)

Cword

Dword.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using word.lower() instead of uppercase.

Using the wrong variable name in the comprehension.

The comprehension maps the uppercase version of each word to its length using len(word). The variable word iterates over the list words. The condition filters words longer than 3 characters.