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

Time Complexity: Longest Word in Dictionary Using Trie

O(N * L)

Understanding Time Complexity

We want to understand how the time needed grows when finding the longest word in a dictionary using a Trie.

How does the number of steps change as the dictionary gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct TrieNode {
    children: [Option; 26],
    is_word: bool,
}

impl TrieNode {
    fn new() -> Self {
        TrieNode {
            children: Default::default(),
            is_word: false,
        }
    }

    fn insert(&mut self, word: &str) {
        let mut node = self;
        for ch in word.chars() {
            let idx = (ch as usize) - ('a' as usize);
            if node.children[idx].is_none() {
                node.children[idx] = Some(Box::new(TrieNode::new()));
            }
            node = node.children[idx].as_mut().unwrap();
        }
        node.is_word = true;
    }
}

This code inserts words into a Trie, building a tree where each node represents a letter.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character of every word to insert it into the Trie.
• How many times: For each word, the loop runs once per character; for all words, it runs for the total number of characters in all words.

How Execution Grows With Input

As the number of words and their lengths grow, the total steps increase roughly with the total characters inserted.

Input Size (n words)	Approx. Operations (total characters)
10 words, avg length 5	~50 steps
100 words, avg length 5	~500 steps
1000 words, avg length 5	~5000 steps

Pattern observation: The operations grow linearly with the total number of characters inserted.

Final Time Complexity

Time Complexity: O(N * L)

This means the time grows linearly with the number of words (N) times the average length of each word (L).

Common Mistake

[X] Wrong: "Inserting words into a Trie is always O(1) because it's a tree."

[OK] Correct: Each word requires steps equal to its length, so longer words take more time, making the time depend on word length, not constant.

Interview Connect

Understanding this helps you explain how tries efficiently handle many words and why their insertion time depends on word length, a key insight for many coding problems.

Self-Check

"What if we changed the Trie to store words with uppercase letters as well? 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 their prefixes, allowing quick prefix checks.

2. Step 2: Relate to longest word search

   Finding the longest word with all prefixes present is easier with Trie traversal.

3. Final Answer:

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

4. Quick Check:

   Trie = prefix-based search [OK]

Hint: Trie is all about prefixes and fast prefix lookups [OK]

Common Mistakes:

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

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

easy

A. struct TrieNode { children: [Option; 26] }

B. struct TrieNode { children: HashMap }

C. struct TrieNode { children: Vec> }

D. struct TrieNode { children: Vec<TrieNode> }

Solution

1. Step 1: Understand Trie children representation

   Each node has fixed 26 children for letters a-z, stored as array of Options.

2. Step 2: Check Rust syntax correctness

   Option allows nullable child nodes; array size 26 matches alphabet.

3. Final Answer:

   struct TrieNode { children: [Option; 26] } -> Option A

4. Quick Check:

   Fixed-size array with Option<Box> = correct Trie node [OK]

Hint: Use fixed-size array with Option<Box> for children in Rust Trie [OK]

Common Mistakes:

• Using Vec without fixed size wastes memory
• Using HashMap without Box causes ownership errors
• Not wrapping children in Option causes panic

3. Given the following Rust code snippet that inserts words into a Trie and finds the longest word with all prefixes present, what will be the printed output?

let words = vec!["a", "app", "ap", "apply", "apple"];
let longest = find_longest_word(words);
println!("{}", longest);

Assume find_longest_word uses a Trie to find the longest word where every prefix is also a word.

medium

A. "apple"

B. "apply"

C. "app"

D. "ap"

Solution

1. Step 1: Check prefixes for each word

   "apple" prefixes: a, ap, app, appl; "appl" missing, so invalid.
   "apply" prefixes: a, ap, app, appl; "appl" missing, invalid.
   "app" prefixes: a, ap, app; all present.
   "ap" prefixes: a, ap; all present.

2. Step 2: Find longest valid word

   "app" is longest word with all prefixes present.

3. Final Answer:

   "app" -> Option C

4. Quick Check:

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

Hint: Check all prefixes exist before choosing longest word [OK]

Common Mistakes:

• Assuming 'apple' or 'apply' are valid without all prefixes
• Ignoring prefix 'appl' missing
• Choosing shortest word instead of longest

4. The following Rust function attempts to insert words into a Trie and find the longest word with all prefixes present. It compiles but returns an empty string. What is the likely bug?

fn find_longest_word(words: Vec<&str>) -> String {
    let mut trie = Trie::new();
    for word in &words {
        trie.insert(word);
    }
    trie.search_longest_word()
}

// Trie methods:
// insert(&self, word: &str)
// search_longest_word(&self) -> String

medium

A. The function should return &str, not String

B. The insert method should take &mut self, not &self

C. Words vector should be mutable

D. The search_longest_word method should take &mut self, not &self

Solution

1. Step 1: Analyze method signatures

   Inserting words modifies Trie, so insert must take mutable reference &mut self.

2. Step 2: Identify compilation vs runtime issue

   Code compiles but insert called with &self means no actual insertion happens, so Trie remains empty.

3. Final Answer:

   The insert method should take &mut self, not &self -> Option B

4. Quick Check:

   Mutable insert needed for Trie modification [OK]

Hint: Insert modifies Trie; needs &mut self in Rust [OK]

Common Mistakes:

• Thinking search needs mutable reference
• Making words vector mutable unnecessarily
• Returning wrong type String vs &str

5. You have a list of words: ["a", "banana", "app", "appl", "ap", "apply", "apple"]. Using a Trie, which word will be returned as the longest word where every prefix is also a word? Explain why the word "banana" is not chosen.

hard

A. "apple" because "banana" prefixes are missing

B. "apply" because it is longer than "apple"

C. "appl" because "appl" is a prefix of "apple" and "apply"

D. "apple" because all its prefixes exist

Solution

1. Step 1: Check prefixes of each candidate word

   "apple" prefixes: a, ap, app, appl, apple - all present.
   "apply" prefixes: a, ap, app, appl, apply - all present.
   "banana" prefixes: b, ba, ban, bana, banan, banana - none except 'b' likely missing.

2. Step 2: Understand why "banana" is excluded

   Since prefixes like 'b', 'ba' are not in the word list, "banana" is invalid.
   Between "apple" and "apply", both valid, but "apple" is chosen (depends on lex order or implementation).

3. Final Answer:

   "apple" because "banana" prefixes are missing -> Option A

4. Quick Check:

   Missing prefixes exclude "banana" [OK]

Hint: Check all prefixes exist; missing prefixes exclude word [OK]

Common Mistakes:

• Assuming "banana" is valid without prefixes
• Confusing "apply" as longer than "apple"
• Ignoring prefix checks for all words