Count Words with Given Prefix in DSA C++ - Execution Trace

Concept Flow - Count Words with Given Prefix

Start at root node

↓

For each char in prefix

↓

Check if char child exists?

No→Prefix not found, return 0

Yes↓

Move to child node

↓

Repeat for next char

↓

After last char

↓

Return count of words with this prefix

Start from the root of the trie and follow each character of the prefix. If any character is missing, return 0. Otherwise, after the last character, return the count of words that share this prefix.

Execution Sample

DSA C++

class TrieNode {
public:
  TrieNode* children[26];
  int prefixCount;
  TrieNode() : prefixCount(0) {
    for (int i = 0; i < 26; i++) children[i] = nullptr;
  }
};

int countWordsWithPrefix(TrieNode* root, const string& prefix) {
  TrieNode* curr = root;
  for (char c : prefix) {
    int idx = c - 'a';
    if (!curr->children[idx]) return 0;
    curr = curr->children[idx];
  }
  return curr->prefixCount;
}

This code counts how many words in the trie start with the given prefix by traversing nodes for each prefix character and returning the stored prefix count.

Execution Table

Step	Operation	Current Node	Character	Child Exists?	Action	Prefix Count	Visual State
1	Start at root	root	-	N/A	Move to next char	5	root(5)
2	Check 'a'	root	a	Yes	Move to child 'a'	3	root(5) -> a(3)
3	Check 'p'	a	p	Yes	Move to child 'p'	3	root(5) -> a(3) -> p(3)
4	Check 'p'	p	p	Yes	Move to child 'p'	2	root(5) -> a(3) -> p(3) -> p(2)
5	Check 'l'	p	l	Yes	Move to child 'l'	2	root(5) -> a(3) -> p(3) -> p(2) -> l(2)
6	Check 'e'	l	e	Yes	Move to child 'e'	1	root(5) -> a(3) -> p(3) -> p(2) -> l(2) -> e(1)
7	Prefix end	e	-	N/A	Return prefixCount	1	root(5) -> a(3) -> p(3) -> p(2) -> l(2) -> e(1)

💡 Traversal ends after last prefix character 'e'; prefixCount 1 returned.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
curr	root	node 'a'	node 'p'	node 'p'	node 'l'	node 'e'	node 'e'
prefixCount	-	3	3	2	2	1	1

Key Moments - 3 Insights

Why do we return 0 if a child node for a prefix character does not exist?

Because if any character in the prefix is missing in the trie, it means no word starts with that prefix, so the count is zero.

Why do we return prefixCount at the last node of the prefix?

The prefixCount at the last node stores how many words have that prefix (see Step 7 in execution_table), so returning it gives the total count of words starting with the prefix.

What does prefixCount represent at each node?

prefixCount counts how many words pass through that node, meaning how many words have the prefix up to that character (visible in variable_tracker and execution_table).

Visual Quiz - 3 Questions

Look at the execution_table at Step 4, what is the current node and character checked?

A Node 'p', character 'p'

B Node 'a', character 'p'

C Node 'p', character 'l'

D Node 'root', character 'a'

💡 Hint

Check the 'Current Node' and 'Character' columns in execution_table row for Step 4.

At which step does the traversal reach the last character of the prefix?

A Step 5

B Step 7

C Step 6

D Step 4

💡 Hint

Look for the step where the last prefix character 'e' is checked in execution_table.

If the prefix was 'appx' instead of 'apple', what would happen in the execution_table?

A Traversal would return prefixCount at node 'p'

B Traversal would return 0 at the missing child for 'x'

C Traversal would continue normally and return prefixCount at 'e'

D Traversal would skip the missing character and continue

💡 Hint

Refer to the 'Child Exists?' and 'Action' columns for missing child cases in execution_table.

Concept Snapshot

Count Words with Given Prefix in Trie:
- Start at root node
- For each prefix char, move to child node
- If child missing, return 0
- After last char, return prefixCount stored
- prefixCount = number of words sharing that prefix

Full Transcript

To count words with a given prefix in a trie, start at the root node and follow each character of the prefix. If at any point the child node for a character does not exist, return zero because no words have that prefix. If all characters exist, after the last character, return the prefixCount stored at that node. This count tells how many words start with the prefix. The code traverses nodes for each prefix character and returns the stored count. The execution table shows each step moving through nodes for characters 'a', 'p', 'p', 'l', 'e' and finally returning the count 1. Variables track the current node and prefixCount at each step. Key moments clarify why missing children return zero and why prefixCount is returned at the last node. The visual quiz tests understanding of node traversal and missing child handling.