Longest Word in Dictionary Using Trie in DSA Rust - Execution Trace

Concept Flow - Longest Word in Dictionary Using Trie

Build Trie from words

↓

Insert each word letter by letter

↓

Mark end of word nodes

↓

Traverse Trie DFS

↓

Check if node is end of word

↓

Update longest word if longer

↓

Return longest word found

Build a Trie from the dictionary words, then traverse it depth-first to find the longest word where all prefixes are valid words.

Execution Sample

DSA Rust

struct TrieNode {
    children: [Option<Box<TrieNode>>; 26],
    is_end: bool,
}

// Insert words and find longest word with all prefixes
fn longest_word(words: Vec<&str>) -> String { ... }

Builds a Trie from given words and finds the longest word where every prefix is also a word in the dictionary.

Execution Table

Step	Operation	Nodes in Trie	Pointer Changes	Visual State
1	Insert 'a'	Root -> a (is_end: true)	root.children['a'] = new node	root -> a*
2	Insert 'app'	Root -> a* -> p -> p (is_end: true)	a.children['p'] = new node, p.children['p'] = new node	root -> a* -> p -> p*
3	Insert 'apple'	Root -> a* -> p -> p* -> l -> e (is_end: true)	p.children['l'] = new node, l.children['e'] = new node	root -> a* -> p -> p* -> l -> e*
4	Insert 'apply'	Root -> a* -> p -> p* -> l -> e*, y (is_end: true)	e.children['y'] = new node	root -> a* -> p -> p* -> l -> e*, y*
5	Insert 'banana'	Root -> a* ... b -> a -> n -> a -> n -> a (is_end: true)	root.children['b'] = new node, chain for 'banana'	root -> a* ... b* -> a -> n -> a -> n -> a*
6	DFS Traverse from root	Traverse nodes where is_end == true	Move pointer down valid paths	Check prefixes: a, app, apple, apply, banana
7	Update longest word	Longest word updated to 'apple'	Compare lengths and prefix validity	Longest word = 'apple'
8	Finish traversal	Return longest word	No more nodes to visit	Result: 'apple'

💡 Traversal ends after all nodes with valid prefixes are visited.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
Trie Size (nodes)	1 (root)	2	4	6	7	13	13	13
Longest Word	""	"a"	"app"	"apple"	"apple"	"apple"	"apple"	"apple"
Current Node	root	a	p	p	l	b	DFS traversal	end
Pointer Position	root	a	p	p	l	b	DFS moves	none

Key Moments - 3 Insights

Why do we only consider nodes where is_end is true during DFS?

Because the problem requires all prefixes to be valid words. The execution_table rows 6 and 7 show traversal only continues down nodes marked as word ends.

How does the Trie structure help find the longest word efficiently?

The Trie stores all words sharing prefixes in a tree. This lets us check prefix validity by node is_end flags during traversal, as seen in execution_table steps 1-5 building the Trie.

Why is the longest word updated at step 7 and not earlier?

Because only after traversing nodes with valid prefixes can we confirm the word is valid. Step 7 compares lengths and updates longest word accordingly.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the visual state of the Trie?

A root -> a* -> p -> p* -> l -> e*

B root -> a* -> p -> p -> l

C root -> a* -> p -> p*

D root -> a* -> p -> l -> e*

💡 Hint

Refer to the Visual State column at step 3 in execution_table.

At which step does the Trie size reach 13 nodes?

A After Step 4

B After Step 5

C After Step 6

D After Step 7

💡 Hint

Check the Trie Size row in variable_tracker after each step.

If the word 'app' was missing, how would the longest word change?

A Longest word would be 'apple'

B Longest word would be 'apply'

C Longest word would be 'a'

D Longest word would be 'banana'

💡 Hint

Consider that 'app' is a prefix needed for 'apple' and 'apply' to be valid, see key_moments explanation.

Concept Snapshot

Longest Word in Dictionary Using Trie
- Build Trie inserting each word letter by letter
- Mark nodes where words end
- DFS traverse only nodes with is_end true
- Update longest word if current path is longer
- Return longest word with all prefixes valid

Full Transcript

We build a Trie from the given dictionary words by inserting each letter as a node. Each node marks if it ends a valid word. Then we traverse the Trie depth-first, only continuing down nodes that mark the end of a word to ensure all prefixes are valid. During traversal, we track the longest word found. Finally, we return the longest word where every prefix is also a word in the dictionary.

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