Trie Insert Operation in DSA C++ - Time & Space Complexity

Time Complexity: Trie Insert Operation

O(n)

Understanding Time Complexity

We want to understand how the time to insert a word into a trie changes as the word gets longer.

How does the number of steps grow when adding more letters?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct TrieNode {
    TrieNode* children[26] = {nullptr};
    bool isEnd = false;
};

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

This code inserts each letter of a word into the trie, creating new nodes if needed.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop over each character in the word.
• How many times: Exactly once per letter in the word (word length = n).

How Execution Grows With Input

Each letter causes one step to check or create a node, so the steps grow directly with word length.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The time grows linearly as the word gets longer.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert grows in direct proportion to the length of the word.

Common Mistake

[X] Wrong: "Insertion time depends on the number of words already in the trie."

[OK] Correct: The insertion only depends on the length of the word being added, not how many words are stored.

Interview Connect

Knowing this helps you explain why tries are efficient for prefix searches and insertions, a common topic in coding interviews.

Self-Check

"What if we changed the alphabet size from 26 to 256? How would the time complexity change?"

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