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?"

Practice

1. What is the main purpose of using a Trie data structure when finding the longest word in a dictionary?

easy

A. To reverse the words quickly

B. To sort words alphabetically

C. To count the total number of words

D. To efficiently store and search words by their prefixes

Solution

1. Step 1: Understand Trie usage

   A Trie stores words by breaking them into characters and linking nodes for prefixes.

2. Step 2: Connect Trie to longest word search

   This structure allows quick prefix checks, helping find the longest word built from other words.

3. Final Answer:

   To efficiently store and search words by their prefixes -> Option D

4. Quick Check:

   Trie = prefix search [OK]

Hint: Trie is best for prefix-based word searches [OK]

Common Mistakes:

• Confusing Trie with sorting algorithms
• Thinking Trie reverses words
• Assuming Trie counts words only

2. Which of the following Go code snippets correctly initializes a Trie node struct for storing children nodes?

easy

A. type TrieNode struct { children map[int]string; isWord string }

B. type TrieNode struct { children []int; isWord int }

C. type TrieNode struct { children map[rune]*TrieNode; isWord bool }

D. type TrieNode struct { children []string; isWord bool }

Solution

1. Step 1: Identify correct children type

   Children should map characters (rune) to TrieNode pointers for branching.

2. Step 2: Check isWord type

   isWord is a boolean indicating if node ends a valid word.

3. Final Answer:

   type TrieNode struct { children map[rune]*TrieNode; isWord bool } -> Option C

4. Quick Check:

   Children map[rune]*TrieNode and bool isWord [OK]

Hint: Children map uses rune keys and TrieNode pointers [OK]

Common Mistakes:

• Using slices instead of maps for children
• Wrong data types for children or isWord
• Using int or string instead of rune keys

3. Given the following Go code snippet that inserts words into a Trie and finds the longest word, what will be printed?

words := []string{"a", "app", "apple", "apply", "ap"}
trie := NewTrie()
for _, w := range words {
    trie.Insert(w)
}
result := trie.LongestWord()
fmt.Println(result)

medium

A. "app"

B. "apply"

C. "a"

D. "apple"

Solution

1. Step 1: Insert words into Trie

   All words "a", "ap", "app", "apple", "apply" are inserted.

2. Step 2: Find longest word built from prefixes

   "app" has all prefixes "a", "ap", "app" present as words. "apple" and "apply" lack "appl" prefix.

3. Final Answer:

   "app" -> Option A

4. Quick Check:

   Longest word with all prefixes = "app" [OK]

Hint: Longest word with all prefixes present wins, lex order breaks ties [OK]

Common Mistakes:

• Choosing "apple" or "apply" ignoring missing "appl"
• Picking lexicographically later word
• Not inserting all words before search

4. In the following Go code for inserting words into a Trie, what is the bug that prevents correct longest word detection?

func (t *Trie) Insert(word string) {
    node := t.root
    for _, ch := range word {
        if node.children[ch] == nil {
            node.children[ch] = &TrieNode{}
        }
        node = node.children[ch]
    }
    node.isWord = false
}

medium

A. isWord should be set to true, not false

B. children map is not initialized

C. Loop variable ch should be int, not rune

D. Function should return a value

Solution

1. Step 1: Check isWord assignment

   isWord marks end of a valid word and must be true, not false.

2. Step 2: Understand impact on longest word search

   Setting isWord false means no word end is recognized, breaking longest word logic.

3. Final Answer:

   isWord should be set to true, not false -> Option A

4. Quick Check:

   isWord = true marks word end [OK]

Hint: Set isWord true at word end to mark valid words [OK]

Common Mistakes:

• Setting isWord false instead of true
• Not initializing children map before use
• Ignoring rune type for characters

5. You have a list of words: ["w", "wo", "wor", "worl", "world", "banana", "ban", "bana", "banan"]. Using a Trie, which word will be the longest word in the dictionary where every prefix is also a word? Assume ties are broken by lexicographical order.

hard

A. "worl"

B. "world"

C. "banana"

D. "banan"

Solution

1. Step 1: Check prefixes for each candidate

   "world" has prefixes "w", "wo", "wor", "worl" all present.

2. Step 2: Check "banana" and its prefixes

   "banana" lacks prefixes like "b", "ba", so it fails.

3. Step 3: Compare longest valid words

   "world" (length 5) is longest with all prefixes present; "banana" (6) invalid, "banan" also invalid.

4. Final Answer:

   "world" -> Option B

5. Quick Check:

   Longest word with all prefixes = "world" [OK]

Hint: Check all prefixes exist; longest with lex order wins [OK]

Common Mistakes:

• Ignoring missing prefixes for banana
• Choosing shorter words
• Not applying lexicographical tie-break