Longest Word in Dictionary Using Trie in DSA Go - Time & Space Complexity

Time Complexity: Longest Word in Dictionary Using Trie

O(n * m)

Understanding Time Complexity

We want to understand how the time needed to find the longest word in a dictionary using a Trie grows as the dictionary size increases.

Specifically, how does the search and insertion time change when we add more words?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Insert words into Trie
for _, word := range words {
    node := root
    for _, ch := range word {
        if node.children[ch-'a'] == nil {
            node.children[ch-'a'] = &TrieNode{}
        }
        node = node.children[ch-'a']
    }
    node.isWord = true
}

// Find longest word by DFS
func dfs(node *TrieNode, path string) string {
    longest := path
    for i, child := range node.children {
        if child != nil && child.isWord {
            candidate := dfs(child, path+string('a'+i))
            if len(candidate) > len(longest) || (len(candidate) == len(longest) && candidate < longest) {
                longest = candidate
            }
        }
    }
    return longest
}

This code inserts all words into a Trie and then uses depth-first search to find the longest word where every prefix is also a word.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Insertion loops over each character of every word, and DFS recursively explores Trie nodes.
• How many times: Insertion runs once per word and per character; DFS visits nodes that represent valid words.

How Execution Grows With Input

As the number of words (n) and their average length (m) grow, insertion takes longer because each character is processed.

Input Size (n)	Approx. Operations
10 words, avg length 5	~50 insert steps + DFS visits
100 words, avg length 5	~500 insert steps + DFS visits
1000 words, avg length 5	~5000 insert steps + DFS visits

Pattern observation: Operations grow roughly proportional to total characters inserted and nodes visited during DFS.

Final Time Complexity

Time Complexity: O(n * m)

This means the time grows linearly with the total number of characters in all words combined.

Common Mistake

[X] Wrong: "The search for the longest word is just O(n) because we only check each word once."

[OK] Correct: Each word insertion processes all its characters, and DFS explores nodes based on prefixes, so the total work depends on all characters, not just word count.

Interview Connect

Understanding how Trie operations scale helps you explain efficient prefix searches and word validations, a useful skill in many coding challenges.

Self-Check

"What if we used a hash set instead of a Trie for prefix checks? How would the time complexity change?"