Why Count Words with Given Prefix in DSA C++?

The Big Idea

What if you could find all words starting with your favorite letters instantly, no matter how big the list?

The Scenario

Imagine you have a huge list of words, like a dictionary, and you want to find out how many words start with a certain set of letters, like "pre".

Doing this by checking each word one by one is like looking for all your friends in a crowd by calling their full names loudly and waiting for a response.

The Problem

Checking each word manually is very slow and tiring, especially if the list is very long.

You might miss some words or make mistakes counting them.

It's like trying to find all your friends in a stadium by shouting their names one by one--it takes too long and is easy to mess up.

The Solution

Using a smart structure like a prefix tree (Trie) helps you quickly find how many words start with the letters you want.

It organizes words by their letters, so you can jump straight to the part where words with your prefix live and count them fast.

Before vs After

✗ Before

int countWordsWithPrefix(vector<string> words, string prefix) {
    int count = 0;
    for (string word : words) {
        if (word.substr(0, prefix.size()) == prefix) {
            count++;
        }
    }
    return count;
}

✓ After

struct TrieNode {
    TrieNode* children[26] = {nullptr};
    int prefixCount = 0;
};

class Trie {
    TrieNode* root = new TrieNode();
public:
    void insert(string word) {
        TrieNode* current = root;
        for (char ch : word) {
            int index = ch - 'a';
            if (!current->children[index]) {
                current->children[index] = new TrieNode();
            }
            current = current->children[index];
            current->prefixCount++;
        }
    }
    int countWordsWithPrefix(string prefix) {
        TrieNode* current = root;
        for (char ch : prefix) {
            int index = ch - 'a';
            if (!current->children[index]) return 0;
            current = current->children[index];
        }
        return current->prefixCount;
    }
};

What It Enables

This lets you instantly know how many words start with any prefix, even in huge word lists, making searches lightning fast.

Real Life Example

Search engines use this to suggest words as you type, showing you popular words starting with the letters you entered.

Key Takeaways

Manual checking is slow and error-prone for large word lists.

Using a Trie organizes words by letters for fast prefix counting.

This method speeds up searches and suggestions in real applications.

Practice

1. What does the problem "Count Words with Given Prefix" ask you to do?

easy

A. Count how many words in a list start with a specific prefix

B. Count the total number of words in a list

C. Find the longest word in a list

D. Replace the prefix in all words with another string

Solution

1. Step 1: Understand the problem statement

   The problem asks to find how many words start with a given prefix from a list of words.

2. Step 2: Identify the key operation

   We need to check each word and count those that begin with the prefix.

3. Final Answer:

   Count how many words in a list start with a specific prefix -> Option A

4. Quick Check:

   Counting words with prefix = Count how many words in a list start with a specific prefix [OK]

Hint: Focus on words starting with prefix, not total words [OK]

Common Mistakes:

• Counting all words instead of only those with prefix
• Looking for longest word instead of counting
• Replacing prefix instead of counting

2. Which of the following C++ code snippets correctly checks if a string word starts with prefix pref?

easy

A. if (word.compare(pref) == 1) { /* starts with prefix */ }

B. if (word == pref) { /* starts with prefix */ }

C. if (word.substr(1, pref.size()) == pref) { /* starts with prefix */ }

D. if (word.find(pref) == 0) { /* starts with prefix */ }

Solution

1. Step 1: Understand string::find usage

   Using word.find(pref) == 0 checks if pref is found at the start (index 0) of word.

2. Step 2: Check other options

   if (word == pref) { /* starts with prefix */ } checks equality, not prefix. if (word.substr(1, pref.size()) == pref) { /* starts with prefix */ } checks substring starting at index 1, which is wrong. if (word.compare(pref) == 1) { /* starts with prefix */ } compares strings but result 1 means greater, not prefix.

3. Final Answer:

   if (word.find(pref) == 0) { /* starts with prefix */ } -> Option D

4. Quick Check:

   find(pref) == 0 means prefix match [OK]

Hint: Use find(pref) == 0 to check prefix in C++ strings [OK]

Common Mistakes:

• Using equality instead of prefix check
• Checking substring from wrong index
• Misunderstanding compare() return values

3. What is the output of the following C++ code?

std::vector words = {"apple", "app", "application", "banana"};
std::string prefix = "app";
int count = 0;
for (const auto& w : words) {
    if (w.find(prefix) == 0) {
        count++;
    }
}
std::cout << count << std::endl;

medium

A. 3

B. 2

C. 1

D. 4

Solution

1. Step 1: Check each word for prefix "app"

   "apple" starts with "app" (count=1), "app" starts with "app" (count=2), "application" starts with "app" (count=3), "banana" does not.

2. Step 2: Count total matches

   Total words starting with "app" are 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Words starting with "app" = 3 [OK]

Hint: Count words where find(prefix) == 0 inside loop [OK]

Common Mistakes:

• Counting words that contain prefix anywhere
• Off-by-one errors in counting
• Ignoring words that exactly equal prefix

4. Identify the error in this C++ code snippet that tries to count words starting with prefix pref:

int count = 0;
for (auto w : words) {
    if (w.find(pref)) {
        count++;
    }
}

medium

A. The count variable should be initialized to -1

B. The loop variable should be a reference: auto& w

C. The condition should check if w.find(pref) == 0, not just w.find(pref)

D. The find function should be replaced with substr

Solution

1. Step 1: Understand find() return value

   find() returns position or npos. Using if(w.find(pref)) treats 0 as false, so words starting at index 0 are skipped.

2. Step 2: Correct condition for prefix check

   We must check if w.find(pref) == 0 to confirm prefix at start.

3. Final Answer:

   The condition should check if w.find(pref) == 0, not just w.find(pref) -> Option C

4. Quick Check:

   Check find() == 0 for prefix [OK]

Hint: Always compare find() result to 0 for prefix check [OK]

Common Mistakes:

• Using if(w.find(pref)) which fails for prefix at start
• Not using reference in loop (less critical here)
• Initializing count incorrectly

5. Given a list of words and a prefix, how would you efficiently count words starting with the prefix if the list is very large and queried multiple times?
Choose the best approach.

hard

A. Check each word with find(prefix) == 0 every time a query comes

B. Use a Trie (prefix tree) to store words and count prefix matches quickly

C. Sort the list and use binary search to find prefix range each time

D. Store words in a hash set and check prefix membership

Solution

1. Step 1: Understand repeated prefix queries

   Checking each word every time is slow for large lists and many queries.

2. Step 2: Use Trie data structure

   A Trie stores words by prefixes, allowing fast counting of words with given prefix in O(length of prefix) time.

3. Step 3: Evaluate other options

   Binary search after sorting helps but is less efficient for many queries. Hash set doesn't support prefix queries efficiently.

4. Final Answer:

   Use a Trie (prefix tree) to store words and count prefix matches quickly -> Option B

5. Quick Check:

   Trie = fast prefix counting [OK]

Hint: Use Trie for fast repeated prefix counts [OK]

Common Mistakes:

• Checking all words each query (slow)
• Using hash set which doesn't support prefix queries
• Ignoring Trie advantages for prefix problems