Longest Word in Dictionary Using Trie in DSA Go - 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

↓

For each child node

↓

Check if node is end of a word

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Update longest word if longer

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Return longest word found

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

Execution Sample

DSA Go

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

Insert words into trie and find the longest word with all prefixes present.

Execution Table

Step	Operation	Nodes in Trie	Pointer Changes	Visual State
1	Insert 'a'	root -> a(end)	root.children['a'] created	root -> a(end) a(end)
2	Insert 'app'	root -> a(end) -> p -> p(end)	a.children['p'] created, p.children['p'] created, last p marked end	root -> a(end) a(end) -> p p -> p(end)
3	Insert 'apple'	root -> a(end) -> p -> p(end) -> l -> e(end)	p.children['l'] created, l.children['e'] created, e marked end	root -> a(end) a(end) -> p p -> p(end) -> l l -> e(end)
4	Insert 'apply'	root -> a(end) -> p -> p(end) -> l -> e(end), y(end)	l.children['y'] created, y marked end	root -> a(end) a(end) -> p p -> p(end) -> l l -> e(end), y(end)
5	Insert 'ap'	root -> a(end) -> p(end) -> p(end) -> l -> e(end), y(end)	a.children['p'] marked end	root -> a(end) a(end) -> p(end) p -> p(end) -> l l -> e(end), y(end)
6	Start DFS from root	No change	current = root	root
7	DFS to 'a' node (end=true)	No change	current = a	root -> a(end)
8	DFS to 'p' node (end=true)	No change	current = p	root -> a(end) -> p(end)
9	DFS to next 'p' node (end=true)	No change	current = p	root -> a(end) -> p(end) -> p(end)
10	DFS to 'l' node (end=false)	Skip this branch	No pointer change	Skip l branch since end=false
11	DFS to 'l' node (end=false) skipped, backtrack	No change	Back to p node	root -> a(end) -> p(end) -> p(end)
12	Update longest word to 'app'	No change	longest = 'app'	Longest word: 'app'
13	DFS to 'p' node (end=true) backtrack to 'a'	No change	Back to a node	root -> a(end)
14	DFS to 'p' node (end=true) backtrack to root	No change	Back to root	root
15	Return longest word 'app'	No change	Done	Longest word: 'app'

💡 DFS ends after exploring all nodes with end=true; longest word found is 'app'

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 12	Final
longest	""	""	""	""	""	""	"app"	"app"
current_node	root	a	p	p	p	p	p	root
Trie Size (nodes)	1	2	4	7	9	9	9	9

Key Moments - 3 Insights

Why do we skip DFS branches where the node is not marked as end of a word?

Because the problem requires all prefixes of the word to be present. If a node is not an end of a word, that prefix is missing. See execution_table step 10 where the 'l' node is skipped.

Why do we mark nodes as end of word during insertion?

Marking nodes as end of word helps us know which prefixes are valid words. This is crucial during DFS to decide if we can continue exploring deeper. Refer to execution_table steps 1-5.

Why does the longest word update only at nodes marked as end of word?

Because only words with all prefixes present count. We update longest word when we reach a node that marks a complete word. See step 12 where longest word updates to 'app'.

Visual Quiz - 3 Questions

Look at the execution_table at step 5, what change happens to the node 'p' after inserting 'ap'?

A The 'p' node is marked as end of a word

B A new child node is created under 'p'

C The root node is marked as end

D No change happens

💡 Hint

Check the 'Pointer Changes' and 'Visual State' columns at step 5 in execution_table

At which step does the longest word get updated to 'app'?

A Step 8

B Step 10

C Step 12

D Step 15

💡 Hint

Look at the 'Operation' and 'Visual State' columns in execution_table for longest word updates

If the word 'apple' was missing from the input, how would the trie size after step 4 change?

A It would be smaller because 'l' and 'e' nodes wouldn't be created

B It would be larger because more nodes are needed

C It would be the same size

D The trie would be empty

💡 Hint

Check variable_tracker for 'Trie Size (nodes)' after step 4 and consider nodes created for 'apple'

Concept Snapshot

Longest Word in Dictionary Using Trie
- Build trie by inserting words letter by letter
- Mark nodes where words end
- Use DFS to explore nodes only if they mark end of a word
- Track longest word during DFS
- Return longest word with all prefixes present

Full Transcript

We build a trie by inserting each word letter by letter, marking nodes where words end. Then we use depth-first search starting from the root. We only continue DFS on nodes that mark the end of a word, ensuring all prefixes exist. During DFS, we update the longest word found. Finally, we return the longest word where all prefixes are present. This method efficiently finds the longest valid word in the dictionary.