Longest Word in Dictionary Using Trie in DSA C# - Execution Trace

Concept Flow - Longest Word in Dictionary Using Trie

Start with empty Trie

↓

Insert each word character by character

↓

Mark end of word nodes

↓

Traverse Trie from root

↓

At each node, check if it forms a valid word

↓

Keep track of longest valid word found

↓

Return longest word after full traversal

Build a Trie by inserting all words, then traverse it to find the longest word where all prefixes are valid words.

Execution Sample

DSA C#

Insert("w")
Insert("wo")
Insert("wor")
Insert("worl")
Insert("world")
FindLongestWord()

Insert words step-by-step into Trie and then find the longest word with all prefixes present.

Execution Table

Step	Operation	Nodes in Trie	Pointer Changes	Visual State
1	Insert 'w'	root -> w(end=true)	root.children['w'] created	root └─ w*
2	Insert 'wo'	root -> w(end=true) -> o(end=true)	w.children['o'] created	root └─ w* └─ o*
3	Insert 'wor'	root -> w(end=true) -> o(end=true) -> r(end=true)	o.children['r'] created	root └─ w* └─ o* └─ r*
4	Insert 'worl'	root -> w(end=true) -> o(end=true) -> r(end=true) -> l(end=true)	r.children['l'] created	root └─ w* └─ o* └─ r* └─ l*
5	Insert 'world'	root -> w(end=true) -> o(end=true) -> r(end=true) -> l(end=true) -> d(end=true)	l.children['d'] created	root └─ w* └─ o* └─ r* └─ l* └─ d*
6	Traverse Trie to find longest word	Same as above	Traverse nodes w->o->r->l->d checking end flags	Longest word found: 'world'

💡 Traversal ends after visiting all nodes with valid word endings.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6
Trie Size (nodes)	1 (root)	2	3	4	5	6	6
Longest Word	""	"w"	"wo"	"wor"	"worl"	"world"	"world"
Current Node	root	node 'w'	node 'o'	node 'r'	node 'l'	node 'd'	null (end)

Key Moments - 3 Insights

Why do we only consider nodes marked as end of word during traversal?

Because only nodes marked as end of word represent valid words. The execution_table rows 6 show traversal checking end flags to ensure prefixes are valid words.

What happens if a prefix is missing in the Trie?

Traversal stops at the first node that is not marked as end of word, as shown in execution_table row 6 where only nodes with end=true are considered for longest word.

Why do we insert words character by character instead of whole word at once?

Inserting character by character builds the Trie structure node by node, allowing shared prefixes. This is shown in execution_table rows 1-5 where each character creates or uses existing nodes.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, how many nodes are in the Trie?

A 3

B 4

C 5

D 6

💡 Hint

Check the 'Nodes in Trie' column at step 3 in execution_table.

At which step does the Trie first contain the word 'worl'?

A Step 3

B Step 4

C Step 5

D Step 6

💡 Hint

Look at the 'Operation' and 'Visual State' columns in execution_table for when 'l' node is created.

If the word 'wor' was not inserted, what would happen during traversal at step 6?

A Longest word would be 'wo'

B Longest word would be 'worl'

C Traversal would continue normally

D Longest word would be 'world'

💡 Hint

Refer to key_moments about traversal stopping at nodes not marked as end of word.

Concept Snapshot

Build a Trie by inserting words character by character.
Mark nodes where words end.
Traverse Trie checking nodes with end=true.
Keep track of longest word with all prefixes valid.
Return longest such word after traversal.

Full Transcript

We start with an empty Trie and insert each word character by character, creating nodes as needed and marking the end of each word. After building the Trie, we traverse it from the root, checking only nodes marked as end of word to ensure all prefixes are valid. We keep track of the longest word found during traversal. The process stops when no further valid prefixes exist. This method efficiently finds the longest word in the dictionary where all prefixes are also words.

Practice

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

easy

A. To sort words alphabetically

B. To efficiently store and search words by their prefixes

C. To count the total number of words

D. To reverse the words in the dictionary

Solution

1. Step 1: Understand Trie structure

   A Trie stores words by sharing common prefixes, making prefix searches fast.

2. Step 2: Relate to longest word search

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

3. Final Answer:

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

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 is the correct way to declare a Trie node class in C# for storing children nodes?

easy

A. public class TrieNode { public Dictionary Children = new Dictionary(); public bool IsWord = false; }

B. public class TrieNode { public List<char> Children; public bool IsWord; }

C. public class TrieNode { public char[] Children = new char[26]; public bool IsWord; }

D. public class TrieNode { public int ChildrenCount; public bool IsWord; }

Solution

1. Step 1: Check children storage type

   Using a Dictionary allows dynamic children mapping by character.

2. Step 2: Confirm boolean flag for word end

   IsWord marks if node completes a valid word.

3. Final Answer:

   public class TrieNode { public Dictionary Children = new Dictionary(); public bool IsWord = false; } -> Option A

4. Quick Check:

   Dictionary + IsWord = correct TrieNode [OK]

Hint: Use Dictionary for flexible children [OK]

Common Mistakes:

• Using List<char> instead of mapping to nodes
• Using char array without node references
• Missing IsWord flag

3. Given the following code snippet, what will be the output?

var words = new string[] {"a", "app", "ap", "apply", "apple"};
var trie = new Trie();
foreach(var w in words) trie.Insert(w);
var longest = trie.FindLongestWord();
Console.WriteLine(longest);

Assuming FindLongestWord returns the longest word where all prefixes exist in the Trie.

medium

A. "ap"

B. "apply"

C. "apple"

D. "app"

Solution

1. Step 1: Insert all words into Trie

   Words inserted: "a", "app", "ap", "apply", "apple".

2. Step 2: 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 exist.
   "ap" prefixes: "a", "ap" exist but shorter than "app".

3. Final Answer:

   "app" -> Option D

4. Quick Check:

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

Hint: Check all prefixes exist for longest word [OK]

Common Mistakes:

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

4. Identify the bug in this C# Trie insertion method snippet:

public void Insert(string word) {
  var node = root;
  foreach (char c in word) {
    if (!node.Children.ContainsKey(c))
      node.Children[c] = new TrieNode();
    node = node.Children[c];
    node.IsWord = true;
  }
}

medium

A. Setting IsWord = true inside the loop for every character

B. Not checking if root is null

C. Using foreach instead of for loop

D. Not initializing Children dictionary

Solution

1. Step 1: Analyze IsWord assignment

   IsWord should be true only at the end of the word, not for every character node.

2. Step 2: Identify correct placement

   IsWord = true must be set after the loop completes, marking the last node.

3. Final Answer:

   Setting IsWord = true inside the loop for every character -> Option A

4. Quick Check:

   IsWord only at word end [OK]

Hint: Set IsWord true only after full word insertion [OK]

Common Mistakes:

• Marking IsWord true too early
• Ignoring Children initialization
• Confusing loop types

5. You have a list of words: ["w", "wo", "wor", "worl", "world", "word", "worst"]. Using a Trie, which word will be returned as the longest word where all prefixes exist in the dictionary?

hard

A. "worst"

B. "word"

C. "world"

D. "worl"

Solution

1. Step 1: Check prefixes for each candidate

   "world" prefixes: "w", "wo", "wor", "worl", "world" all exist.
   "worst" prefixes: "w", "wo", "wor", "wors" (missing), invalid.
   "word" prefixes: "w", "wo", "wor", "word" all exist but shorter than "world".
   "worl" prefixes: "w", "wo", "wor", "worl" all exist but shorter.

2. Step 2: Choose longest valid word

   "world" is longest with all prefixes present.

3. Final Answer:

   "world" -> Option C

4. Quick Check:

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

Hint: Longest word with all prefixes present wins [OK]

Common Mistakes:

• Choosing 'worst' without 'wors' prefix
• Ignoring prefix checks
• Picking shorter words