Trie Insert Operation in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a character into the children map of the current Trie node.

DSA C++

current->children['a'] = new TrieNode();
current = current->children[[1]];

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

A'a'

B'b'

C'c'

D'd'

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different character than the one inserted.

Forgetting to move current to the new node.

We insert the character 'a' into the children map and then move current to that child node.

2fill in blank

medium

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

DSA C++

if (current->children.find(word[[1]]) == current->children.end()) {
    current->children[word[i]] = new TrieNode();
}

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

Aword

B0

Ci

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using a fixed index like 0 or 1 instead of the loop variable.

Using the whole word instead of a single character.

We check if the character at index i in the word exists in the children map.

3fill in blank

hard

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

DSA C++

current->[1] = true;

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

AisEnd

BendOfWord

CisWord

DwordEnd

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that does not exist in the TrieNode class.

Forgetting to mark the end of the word.

The correct member variable to mark the end of a word is 'isEnd'.

4fill in blank

hard

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

DSA C++

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

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 declaration and condition.

Using a variable name not used inside the loop body.

The loop variable should be 'i' consistently for initialization, condition, and increment.

5fill in blank

hard

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

DSA C++

void insert(string [1]) {
    TrieNode* current = root;
    for (int i = 0; i < [2].size(); i++) {
        if (current->children.find([3][i]) == current->children.end()) {
            current->children[[3][i]] = new TrieNode();
        }
        current = current->children[[3][i]];
    }
    current->isEnd = true;
}

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

Aword

Bstr

Dtext

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inside the function than the parameter.

Forgetting to mark the end of the word.

The parameter name and all references inside the function should be consistent, here 'str'.

Practice

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

easy

A. To count the number of words stored in the Trie

B. To delete a word from the Trie

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

D. To search for a word in the Trie

Solution

1. Step 1: Understand the insert operation

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

2. Step 2: Differentiate from 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 C

4. Quick Check:

   Insert = Add word [OK]

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

Common Mistakes:

• Confusing insert with search or delete
• Thinking insert counts words
• Assuming insert replaces existing words

2. Which of the following is the correct way to declare a Trie node class in C++ for insertion?

easy

A. class TrieNode { public: vector<char> children; bool isEnd; TrieNode() { isEnd = false; } };

B. class TrieNode { public: int children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = 0; isEnd = false; } };

C. class TrieNode { public: TrieNode children[26]; bool isEnd; TrieNode() { isEnd = false; } };

D. class TrieNode { public: TrieNode* children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = nullptr; isEnd = false; } };

Solution

1. Step 1: Check children type for TrieNode

   Children must be pointers to TrieNode to allow dynamic node creation; hence TrieNode* children[26].

2. Step 2: Verify initialization

   Initialize all children pointers to nullptr and isEnd to false in constructor.

3. Final Answer:

   class TrieNode { public: TrieNode* children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = nullptr; isEnd = false; } }; -> Option D

4. Quick Check:

   Children pointers + init nullptr = class TrieNode { public: TrieNode* children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = nullptr; isEnd = false; } }; [OK]

Hint: Children must be pointers, initialized to nullptr [OK]

Common Mistakes:

• Using int array instead of pointers
• Not initializing children to nullptr
• Using vector<char> instead of TrieNode pointers

3. What will be the output of the following code snippet after inserting the words "cat" and "car" into an empty Trie?

Trie trie;
trie.insert("cat");
trie.insert("car");
int count = 0;
for(int i = 0; i < 26; i++) {
  if(trie.root->children[i] != nullptr) count++;
}
std::cout << count << std::endl;

medium

A. 1

B. 2

C. 3

D. 0

Solution

1. Step 1: Insert "cat" and "car" into Trie

   Both words start with 'c', so only one child node for 'c' is created at root.

2. Step 2: Count root children nodes

   Only one child at root for 'c' exists, so count is 1.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Root children count after insert = 1 [OK]

Hint: Count root children after insert; common prefix shares node [OK]

Common Mistakes:

• Counting all characters instead of root children
• Assuming separate root nodes for each word
• Confusing total nodes with root children

4. Identify the error in the following Trie insert function snippet:

void insert(string word) {
  TrieNode* node = root;
  for(char c : word) {
    int index = c - 'a';
    if(node->children[index] == NULL) {
      node->children[index] = new TrieNode();
    }
    node = node->children[index];
  }
  node->isEnd = true;
}

medium

A. No error; the insert function is correct

B. The index calculation should be c - 'A' instead of c - 'a'

C. The root pointer is not initialized before insertion

D. The loop should iterate over indices, not characters

Solution

1. Step 1: Check root initialization assumption

   Assuming root is properly initialized elsewhere, this code correctly uses root pointer.

2. Step 2: Verify index calculation and loop

   Index c - 'a' is correct for lowercase letters; loop over characters is valid.

3. Final Answer:

   No error; the insert function is correct -> Option A

4. Quick Check:

   Code logic matches standard insert [OK]

Hint: Check index and pointer usage carefully [OK]

Common Mistakes:

• Assuming uppercase letters without context
• Thinking loop must use indices
• Forgetting root initialization outside snippet

5. Given a Trie that already contains the words "bat", "ball", and "barn", what will be the effect of inserting the word "bar"?
Choose the correct statement.

hard

A. A new branch is created starting from 'b' node

B. No new nodes are created; only an existing node's isEnd flag is set

C. All existing nodes are duplicated to insert the new word

D. The word "bar" cannot be inserted because "barn" exists

Solution

1. Step 1: Analyze existing Trie structure

   Words "bat", "ball", "barn" share prefix 'ba' and 'bar' prefix exists as part of "barn".

2. Step 2: Effect of inserting "bar"

   Nodes for 'b', 'a', 'r' already exist; only isEnd flag at 'r' node is set to true to mark "bar" as a word.

3. Final Answer:

   No new nodes are created; only an existing node's isEnd flag is set -> Option B

4. Quick Check:

   Insert existing prefix sets isEnd only [OK]

Hint: Insert existing prefix sets isEnd flag only [OK]

Common Mistakes:

• Thinking new nodes are always created
• Assuming duplicates of nodes happen
• Believing insertion fails if prefix exists