Trie Node Design and Initialization in DSA C++ - Time & Space Complexity

Time Complexity: Trie Node Design and Initialization

O(1)

Understanding Time Complexity

We want to understand how much work it takes to create a single node in a trie.

This helps us know how fast the trie can grow as we add words.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class TrieNode {
public:
    TrieNode* children[26];
    bool isEndOfWord;

    TrieNode() {
        for (int i = 0; i < 26; i++) {
            children[i] = nullptr;
        }
        isEndOfWord = false;
    }
};
    

This code creates a trie node with 26 children pointers initialized to null and a flag to mark word end.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that sets each child pointer to nullptr.
• How many times: Exactly 26 times, once for each letter of the alphabet.

How Execution Grows With Input

Creating one node always does the same amount of work, no matter how many words we add later.

Input Size (n)	Approx. Operations
1	26 (initializing children)
10	26 (per node, same work each time)
1000	26 (per node, still same work each time)

Pattern observation: The work to create one node stays constant regardless of input size.

Final Time Complexity

Time Complexity: O(1)

This means creating a trie node takes a fixed amount of time, no matter how many nodes or words exist.

Common Mistake

[X] Wrong: "Initializing children depends on the number of words inserted so far."

[OK] Correct: Each node initializes its children array once, always 26 times, independent of how many words are in the trie.

Interview Connect

Understanding node initialization time helps you explain how tries build up efficiently and why each node costs a fixed setup time.

Self-Check

"What if we changed the children array size from 26 to a dynamic map? How would the time complexity of node initialization change?"