Longest Word in Dictionary Using Trie in DSA Python - 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

↓

Start DFS from root

↓

At each node, check if it ends a word

↓

If yes, explore children nodes

↓

Track longest word found

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Return longest word after DFS completes

Build a trie from the dictionary words, then use depth-first search to find the longest word where all prefixes are also words.

Execution Sample

DSA Python

class TrieNode:
    def __init__(self):
        self.children = {}
        self.end_of_word = False

# Insert words and find longest word

This code builds a trie node structure and inserts words to find the longest valid word.

Execution Table

Step	Operation	Nodes in Trie	Pointer Changes	Visual State
1	Insert 'a'	root -> a(end=False)	root.children['a'] created	root └─ a (end=False)
2	Insert 'app'	root -> a(end=False) -> p(end=False) -> p(end=True)	a.children['p'] created, p.children['p'] created, mark end_of_word at last p	root └─ a (end=False) └─ p (end=False) └─ p (end=True)
3	Insert 'apple'	root -> a -> p -> p(end=True) -> l(end=False) -> e(end=True)	p.children['l'] created, l.children['e'] created, mark end_of_word at e	root └─ a └─ p └─ p (end=True) └─ l └─ e (end=True)
4	Insert 'apply'	root -> a -> p -> p(end=True) -> l -> y(end=True)	l.children['y'] created, mark end_of_word at y	root └─ a └─ p └─ p (end=True) └─ l ├─ e (end=True) └─ y (end=True)
5	Insert 'banana'	root -> a ... and b(end=False) -> a -> n -> a -> n -> a(end=True)	root.children['b'] created, chain created for banana, mark end_of_word at last a	root ├─ a │ └─ p │ └─ p (end=True) │ └─ l │ ├─ e (end=True) │ └─ y (end=True) └─ b └─ a └─ n └─ a └─ n └─ a (end=True)
6	DFS start at root	No change	Set longest_word = ''	No visual change
7	DFS visit 'a' node (end=False)	No change	Skip since end_of_word is False	No visual change
8	DFS visit 'a' -> 'p' (end=False)	No change	Skip since end_of_word is False	No visual change
9	DFS visit 'a' -> 'p' -> 'p' (end=True)	No change	Update longest_word = 'app'	No visual change
10	DFS visit 'a' -> 'p' -> 'p' -> 'l' (end=False)	No change	Skip since end_of_word is False	No visual change
11	DFS visit 'a' -> 'p' -> 'p' -> 'l' -> 'e' (end=True)	No change	Update longest_word = 'apple'	No visual change
12	DFS visit 'a' -> 'p' -> 'p' -> 'l' -> 'y' (end=True)	No change	longest_word remains 'apple' (apple < apply lex order)	No visual change
13	DFS visit 'b' node (end=False)	No change	Skip since end_of_word is False	No visual change
14	DFS visit 'b' -> 'a' (end=False)	No change	Skip since end_of_word is False	No visual change
15	DFS visit 'b' -> 'a' -> 'n' (end=False)	No change	Skip since end_of_word is False	No visual change
16	DFS visit 'b' -> 'a' -> 'n' -> 'a' (end=False)	No change	Skip since end_of_word is False	No visual change
17	DFS visit 'b' -> 'a' -> 'n' -> 'a' -> 'n' (end=False)	No change	Skip since end_of_word is False	No visual change
18	DFS visit 'b' -> 'a' -> 'n' -> 'a' -> 'n' -> 'a' (end=True)	No change	longest_word remains 'apple' (banana longer but prefixes missing)	No visual change
19	DFS complete	No change	Return longest_word = 'apple'	No visual change

💡 DFS ends after visiting all nodes with end_of_word True; longest word found is 'apple'

Variable Tracker

Variable	Start	After Step 9	After Step 11	After Step 12	Final
longest_word	''	'app'	'apple'	'apple'	'apple'
current_node	root	'p' node	'e' node	'y' node	DFS complete
visited_nodes	{}	{'a','p','p'}	{'a','p','p','l','e'}	{'a','p','p','l','y'}	All visited

Key Moments - 3 Insights

Why do we only update longest_word at nodes where end_of_word is True?

Because only nodes marked end_of_word represent complete valid words. See execution_table rows 9, 11, and 12 where longest_word updates only at such nodes.

Why does DFS skip nodes where end_of_word is False?

Skipping ensures we only consider words where all prefixes are valid words. For example, in rows 7 and 8, nodes with end_of_word False are skipped to avoid invalid prefixes.

Why is 'banana' not chosen as longest word even though it is longer?

Because its prefixes are not all marked as words in the trie, so DFS skips them (rows 13-18). Only words with all prefixes valid are considered.

Visual Quiz - 3 Questions

Look at the execution_table at step 11, what is the longest_word?

A 'app'

B 'apple'

C 'apply'

D ''

💡 Hint

Check the 'longest_word' column in variable_tracker after step 11

At which step does the DFS find the first complete word?

A Step 9

B Step 7

C Step 13

D Step 18

💡 Hint

Look at execution_table rows where longest_word updates first

If the word 'app' was not inserted, what would happen to longest_word at step 9?

A It would be 'app'

B It would be 'apple'

C It would be '' (empty)

D It would be 'apply'

💡 Hint

Without 'app', the prefix nodes would not be end_of_word True, so longest_word would not update at step 9

Concept Snapshot

Build a trie from all words.
Mark nodes where words end.
Use DFS to explore nodes only if end_of_word is True.
Track the longest word found where all prefixes are valid.
Return the longest such word after DFS completes.

Full Transcript

We build a trie by inserting each word letter by letter, marking nodes where words end. Then, we perform a depth-first search starting from the root. At each node, we only continue if the node marks the end of a word, ensuring all prefixes are valid words. We keep track of the longest word found during this traversal. Finally, we return the longest word found where all prefixes exist 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 efficiently store and search words by their prefixes

B. To sort words alphabetically

C. To count the total number of words

D. To reverse the words in the dictionary

Solution

1. Step 1: Understand Trie usage

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

2. Step 2: Relate Trie to longest word search

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

3. Final Answer:

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

4. Quick Check:

   Trie = prefix search [OK]

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

Common Mistakes:

• Thinking Trie sorts words alphabetically
• Confusing Trie with simple lists
• Assuming Trie counts words only

2. Which of the following is the correct way to insert a word into a Trie node in Python?

easy

A. node.children[char] = TrieNode() if char not in node.children else node.children[char]

B. node.children.add(char)

C. node.children.append(char)

D. node.children[char] = char

Solution

1. Step 1: Understand Trie children structure

   Trie children are stored in a dictionary mapping characters to TrieNode objects.

2. Step 2: Correct insertion method

   We check if character exists; if not, create new TrieNode and assign it to children dictionary.

3. Final Answer:

   node.children[char] = TrieNode() if char not in node.children else node.children[char] -> Option A

4. Quick Check:

   Insert uses dictionary with TrieNode [OK]

Hint: Use dictionary keys for children, not list or set [OK]

Common Mistakes:

• Using add() or append() on dictionary
• Assigning character instead of TrieNode
• Not checking if child exists before insertion

3. Given the words: ["a", "app", "apple", "apply", "ap"], what is the longest word that can be built one character at a time using the Trie method?

medium

A. "apply"

B. "app"

C. "apple"

D. "ap"

Solution

1. Step 1: Check each word's prefixes

   Words must have all prefixes present: "a", "ap", "app" are present; "apple" and "apply" require "appl" which is missing.

2. Step 2: Identify longest valid word

   "app" is the longest word where all prefixes exist in the list.

3. Final Answer:

   "app" -> Option B

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 "appl"
• Ignoring prefix requirement
• Choosing shortest word instead

4. What is the error in this Trie insertion code snippet for building the longest word?

for word in words:
    node = root
    for char in word:
        if char not in node.children:
            node.children[char] = TrieNode
        node = node.children[char]
    node.is_word = True

medium

A. The root node should be reset after each character

B. The loop should iterate over characters in reverse

C. node.is_word should be set before the inner loop

D. TrieNode should be called with parentheses to create an instance

Solution

1. Step 1: Identify object creation mistake

   TrieNode is a class; to create an object, parentheses () are needed.

2. Step 2: Understand impact of missing parentheses

   Without (), node.children[char] stores the class itself, not an instance, causing errors on further access.

3. Final Answer:

   TrieNode should be called with parentheses to create an instance -> Option D

4. Quick Check:

   Instantiate TrieNode with () [OK]

Hint: Always use () to create class instances in Python [OK]

Common Mistakes:

• Forgetting parentheses when creating objects
• Setting flags in wrong loop place
• Resetting root incorrectly inside loops

5. You have a list of words: ["w", "wo", "wor", "worl", "world", "banana", "ban", "bana", "banan"]. Using a Trie, which word will be returned as the longest word built one character at a time from other words in the list?

hard

A. "worl"

B. "banana"

C. "world"

D. "banan"

Solution

1. Step 1: Verify prefix chains for each candidate

   "world" prefixes: "w", "wo", "wor", "worl", "world" all exist; "banana" and "banan" missing prefixes like "b", "ba"; "worl" is shorter than "world".

2. Step 2: Choose longest word with all prefixes present

   "world" is longest with all prefixes present in the list.

3. Final Answer:

   "world" -> Option C

4. Quick Check:

   Longest word with full prefix chain = "world" [OK]

Hint: Check full prefix chain exists before longest word [OK]

Common Mistakes:

• Choosing "banana" without full prefix chain
• Ignoring prefix requirement for "banan"
• Picking shorter word "worl" over "world"