Count Words with Given Prefix in DSA C++ - Build from Scratch

Count Words with Given Prefix

📖 Scenario: You are building a simple program to help a bookstore find how many book titles start with a certain prefix. This helps the store quickly count books by categories or themes.

🎯 Goal: Create a program that counts how many words in a list start with a given prefix.

📋 What You'll Learn

Create a vector of strings called words with the exact values: "apple", "app", "application", "banana", "band"

Create a string variable called prefix and set it to "app"

Use a for loop with variable word to iterate over words

Inside the loop, use word.compare(0, prefix.size(), prefix) == 0 to check if word starts with prefix

Create an integer variable count to count how many words start with prefix

Print the value of count at the end

💡 Why This Matters

🌍 Real World

Counting words with a specific prefix is useful in search engines, autocomplete features, and organizing data by categories.

💼 Career

This skill helps in software development roles involving text processing, search optimization, and building user-friendly interfaces.

1

Create the list of words

Create a vector of strings called words with these exact entries: "apple", "app", "application", "banana", "band"

DSA C++

#include 
#include 
#include 

int main() {
    // Create vector words with "apple", "app", "application", "banana", "band"
    // Your code here
}

Hint

Use std::vector<std::string> and initialize it with the given words inside curly braces.

2

Set the prefix to search

Create a string variable called prefix and set it to "app"

DSA C++

#include 
#include 
#include 

int main() {
    std::vector words = {"apple", "app", "application", "banana", "band"};
    // Create string prefix and set it to "app"
    // Your code here
}

Hint

Use std::string prefix = "app"; to create the prefix variable.

3

Count words starting with the prefix

Create an integer variable called count and set it to 0. Use a for loop with variable word to iterate over words. Inside the loop, use word.compare(0, prefix.size(), prefix) == 0 to check if word starts with prefix. If yes, increase count by 1.

DSA C++

#include 
#include 
#include 

int main() {
    std::vector words = {"apple", "app", "application", "banana", "band"};
    std::string prefix = "app";
    // Create int count = 0
    // Use for loop: for (const std::string& word : words)
    // Inside loop, check if word starts with prefix using word.compare(0, prefix.size(), prefix) == 0
    // If yes, increment count
    // Your code here
}

Hint

Use a range-based for loop and the compare method of std::string to check the prefix.

4

Print the count result

Print the value of count using std::cout

DSA C++

#include 
#include 
#include 

int main() {
    std::vector words = {"apple", "app", "application", "banana", "band"};
    std::string prefix = "app";
    int count = 0;
    for (const std::string& word : words) {
        if (word.compare(0, prefix.size(), prefix) == 0) {
            count++;
        }
    }
    // Print count
    // Your code here
}

Hint

Use std::cout << count << std::endl; to print the count.

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