Trie Insert Operation in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

if char not in node.children:
    node.children[[1]] = TrieNode()

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

Achar

Bnode

Cchildren

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' or 'children' instead of the character as the key.

Using 'word' which is the whole string, not a single character.

We use 'char' as the key to insert a new TrieNode in the children dictionary.

2fill in blank

medium

Complete the code to move to the next node after inserting or finding the character.

DSA Python

node = node.[1][char]

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

Akeys

Bitems

Cvalues

Dchildren

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'keys', 'values', or 'items' which are dictionary methods, not attributes.

We access the 'children' dictionary of the current node to move to the next node.

3fill in blank

hard

Fix the error in marking the end of a word in the Trie node.

DSA Python

node.[1] = True

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

Ais_end

Bend

Cend_of_word

Dis_word

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'end' or 'end_of_word' which may not be defined in the TrieNode class.

Using 'is_word' which is not the standard attribute name.

The standard attribute to mark the end of a word is 'is_end' set to True.

4fill in blank

hard

Fill both blanks to complete the loop that inserts each character of the word into the Trie.

DSA Python

for [1] in [2]:
    if char not in node.children:
        node.children[char] = TrieNode()

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

Achar

Bword

Cnode

Dchildren

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' or 'children' as loop variables which are not iterable strings.

Using 'char' as the iterable instead of the loop variable.

We loop over each 'char' in the 'word' to insert characters one by one.

5fill in blank

hard

Fill all three blanks to complete the insert function for the Trie.

DSA Python

def insert(self, [1]):
    node = self.root
    for [2] in [3]:
        if [2] not in node.children:
            node.children[[2]] = TrieNode()
        node = node.children[[2]]
    node.is_end = True

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

Aword

Bchar

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' as a parameter or loop variable which is incorrect.

Mixing variable names inconsistently.

The function takes 'word', loops over each 'char' in 'word' to insert characters.

Practice

1. What is the main purpose of the insert operation in a Trie data structure?

easy

A. To delete a word from the Trie

B. To add a new word character by character into the Trie

C. To search for a word in the Trie

D. To count the total words stored in the Trie

Solution

1. Step 1: Understand the insert operation

   The insert operation adds each character of a word into the Trie one by one, creating nodes if they don't exist.

2. Step 2: Compare with other operations

   Deleting removes words, searching checks presence, counting totals words. Insert specifically adds words.

3. Final Answer:

   To add a new word character by character into the Trie -> Option B

4. Quick Check:

   Insert = Add word stepwise [OK]

Hint: Insert means adding word letters step-by-step [OK]

Common Mistakes:

• Confusing insert with search or delete
• Thinking insert counts words
• Assuming insert removes words

2. Which of the following is the correct way to start inserting a word "cat" into an empty Trie in Python?

easy

A. for char in "cat": current = current.children[char] if char in current.children else current.children[char] = TrieNode()

B. for char in "cat": current = current.children[char]

C. for char in "cat": current.children[char] = TrieNode(); current = current.children[char]

D. for char in "cat": if char not in current.children: current.children[char] = TrieNode(); current = current.children[char]

Solution

1. Step 1: Check condition before adding node

   We must check if the character node exists before creating a new one to avoid overwriting.

2. Step 2: Move current pointer correctly

   After ensuring the node exists, move current to that child node for next character insertion.

3. Final Answer:

   for char in "cat": if char not in current.children: current.children[char] = TrieNode(); current = current.children[char] -> Option D

4. Quick Check:

   Check existence then add and move [OK]

Hint: Check if node exists before adding new one [OK]

Common Mistakes:

• Not checking if child node exists before assignment
• Not moving current pointer after insertion
• Overwriting existing nodes

3. Consider the following code snippet for inserting the word "dog" into a Trie. What will be the printed output after insertion?

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end = False

root = TrieNode()

word = "dog"
current = root
for char in word:
    if char not in current.children:
        current.children[char] = TrieNode()
    current = current.children[char]
current.is_end = True

print(list(root.children.keys()))

medium

A. ['d']

B. ['o']

C. ['g']

D. []

Solution

1. Step 1: Insert characters one by one

   The first character 'd' is added as a child of root, so root.children has key 'd'.

2. Step 2: Check what is printed

   Printing root.children.keys() shows the first level keys, which is ['d'] after insertion.

3. Final Answer:

   ['d'] -> Option A

4. Quick Check:

   Root children keys after insert = ['d'] [OK]

Hint: Root children keys show first inserted letters [OK]

Common Mistakes:

• Expecting all letters at root level
• Printing keys of last node instead of root
• Confusing keys with values

4. The following code attempts to insert the word "bat" into a Trie but has a bug. What is the error?

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end = False

root = TrieNode()
word = "bat"
current = root
for char in word:
    current.children[char] = TrieNode()
current = current.children[char]
current.is_end = True

medium

A. The children dictionary is not initialized

B. The is_end flag is set before inserting all characters

C. The current pointer is moved outside the loop, causing nodes to not be chained properly

D. The loop variable 'char' is not defined

Solution

1. Step 1: Analyze loop and pointer movement

   The line moving current pointer is outside the loop, so current only moves after loop ends, not per character.

2. Step 2: Understand effect on insertion

   Nodes for each character are created directly under root (no chaining), and only the last node's is_end is set.

3. Final Answer:

   The current pointer is moved outside the loop, causing nodes to not be chained properly -> Option C

4. Quick Check:

   Pointer must move inside loop to build chain [OK]

Hint: Move current pointer inside loop to link nodes [OK]

Common Mistakes:

• Moving pointer outside loop
• Not creating nodes conditionally
• Setting is_end too early

5. You want to insert multiple words ["apple", "app", "ape"] into a Trie. After insertion, which of the following statements is TRUE about the Trie structure?

hard

A. The node for 'l' after "app" exists only for "apple"

B. The node for 'p' after 'ap' will have is_end = True for both "app" and "ape"

C. Each word creates a completely separate branch from root

D. The root node's is_end flag will be True after insertion

Solution

1. Step 1: Understand shared prefixes in Trie

   Words "apple", "app", and "ape" share prefix "ap"; nodes for 'a' and 'p' are shared.

2. Step 2: Check is_end flags for words

   "app" ends at second 'p' node, so that node's is_end = True; "apple" continues to 'l' and 'e', so 'l' node is not end, but 'e' node is end.

3. Step 3: Analyze options

   The node for 'p' after 'ap' will have is_end = True for both "app" and "ape" is wrong because 'p' node cannot be end for both "app" and "ape"; 'ape' ends at 'e' node. Each word creates a completely separate branch from root is wrong because branches share prefix nodes. The root node's is_end flag will be True after insertion is wrong because root node is not end of any word.

4. Final Answer:

   The node for 'l' after "app" exists only for "apple" -> Option A

5. Quick Check:

   Shared prefix nodes, is_end marks word ends [OK]

Hint: Shared prefixes share nodes; is_end marks word ends only [OK]

Common Mistakes:

• Assuming each word has separate root branch
• Marking is_end on wrong nodes
• Setting root is_end True incorrectly