Frequency Counter Pattern Using Hash Map in DSA C++ - Build from Scratch

Frequency Counter Pattern Using Hash Map

📖 Scenario: You are working in a bookstore. You have a list of books sold today. You want to count how many copies of each book were sold.

🎯 Goal: Build a program that counts the frequency of each book sold using a hash map (dictionary) and prints the result.

📋 What You'll Learn

Create a vector of strings called books_sold with the exact book titles given.

Create an empty unordered_map called frequency to count book sales.

Use a for loop with variable book to iterate over books_sold and update frequency.

Print the frequency of each book in the format: Book Title: count.

💡 Why This Matters

🌍 Real World

Counting how many times each item appears is useful in sales analysis, inventory management, and data summarization.

💼 Career

Frequency counting with hash maps is a common technique in software development, data analysis, and coding interviews.

1

Create the list of books sold

Create a vector of strings called books_sold with these exact entries: "Harry Potter", "The Hobbit", "Harry Potter", "Game of Thrones", "The Hobbit", "Harry Potter".

DSA C++

#include 
#include 
#include 

int main() {
    // Create vector books_sold with given book titles
    // Your code here
}

Hint

Use std::vector<std::string> and initialize it with the exact book titles inside curly braces.

2

Create the frequency map

Add an empty unordered_map called frequency that maps std::string to int to count how many times each book appears.

DSA C++

#include 
#include 
#include 
#include 

int main() {
    std::vector books_sold = {"Harry Potter", "The Hobbit", "Harry Potter", "Game of Thrones", "The Hobbit", "Harry Potter"};
    // Create unordered_map frequency to count books
    // Your code here
}

Hint

Include <unordered_map> and declare frequency as std::unordered_map<std::string, int>.

3

Count the frequency of each book

Use a for loop with variable book to iterate over books_sold and increase the count in frequency[book] by 1 for each book.

DSA C++

#include 
#include 
#include 
#include 

int main() {
    std::vector books_sold = {"Harry Potter", "The Hobbit", "Harry Potter", "Game of Thrones", "The Hobbit", "Harry Potter"};
    std::unordered_map frequency;
    // Use for loop to count frequency of each book
    // Your code here
}

Hint

Use a range-based for loop and increment frequency[book] inside the loop.

4

Print the frequency of each book

Use a for loop with variables book and count to iterate over frequency and print each book and its count in the format: Book Title: count.

DSA C++

#include 
#include 
#include 
#include 

int main() {
    std::vector books_sold = {"Harry Potter", "The Hobbit", "Harry Potter", "Game of Thrones", "The Hobbit", "Harry Potter"};
    std::unordered_map frequency;
    for (const std::string& book : books_sold) {
        frequency[book]++;
    }
    // Print each book and its frequency
    // Your code here
}

Hint

Use a range-based for loop with structured bindings to print each book and its count.

Practice

1. What is the main purpose of using a hash map in the Frequency Counter Pattern?

easy

A. To reverse the order of elements

B. To sort the elements in ascending order

C. To store elements without any duplicates

D. To count how many times each element appears in a collection

Solution

1. Step 1: Understand the role of a hash map

   A hash map stores key-value pairs, allowing quick lookup and update of counts.

2. Step 2: Apply to frequency counting

   We use the hash map keys as elements and values as their counts to track frequency.

3. Final Answer:

   To count how many times each element appears in a collection -> Option D

4. Quick Check:

   Frequency Counter = Counting elements [OK]

Hint: Hash map keys track elements, values track counts [OK]

Common Mistakes:

• Thinking hash map sorts elements
• Confusing frequency counting with removing duplicates
• Using hash map to reverse order

2. Which of the following is the correct syntax to declare a hash map in C++ to count frequencies of characters?

easy

A. std::vector<int> freq;

B. std::unordered_map freq;

C. std::unordered_set<char> freq;

D. std::map freq;

Solution

1. Step 1: Identify correct container for frequency counting

   We need a map from characters to integers to store counts.

2. Step 2: Check syntax correctness

   std::unordered_map freq; declares a hash map with char keys and int values.

3. Final Answer:

   std::unordered_map freq; -> Option B

4. Quick Check:

   unordered_map = frequency map [OK]

Hint: Use unordered_map for frequency counting [OK]

Common Mistakes:

• Using map with wrong key-value types
• Using unordered_set which stores only keys
• Using vector which is not a map

3. What will be the output of the following C++ code?

std::string s = "aabbcc";
std::unordered_map freq;
for (char c : s) {
    freq[c]++;
}
for (auto &p : freq) {
    std::cout << p.first << ":" << p.second << " ";
}

medium

A. a:2 b:2 c:2

B. a:1 b:1 c:1

C. a:3 b:3 c:3

D. a:2 b:3 c:1

Solution

1. Step 1: Count frequencies of each character in string

   Characters 'a', 'b', and 'c' each appear twice in "aabbcc".

2. Step 2: Print each character and its count

   The loop prints each character and its frequency, so output is "a:2 b:2 c:2 ".

3. Final Answer:

   a:2 b:2 c:2 -> Option A

4. Quick Check:

   Count 'a','b','c' = 2 each [OK]

Hint: Count each char's occurrences in string [OK]

Common Mistakes:

• Assuming counts are 1 instead of 2
• Mixing counts between characters
• Confusing output format

4. Identify the error in this C++ code snippet that tries to count frequencies of integers in a vector:

std::vector<int> nums = {1, 2, 2, 3};
std::unordered_map freq;
for (int i = 0; i < nums.size(); i++) {
    freq[nums[i]] = freq[nums[i]] + 1;
}
for (auto p : freq) {
    std::cout << p.first << ":" << p.second << " ";
}

medium

A. Loop should use range-based for instead of index

B. Using freq[nums[i]] = freq[nums[i]] + 1; causes runtime error

C. No error; code works correctly

D. freq should be declared as std::unordered_set<int>

Solution

1. Step 1: Check frequency counting logic

   freq[nums[i]] = freq[nums[i]] + 1; correctly increments count for each element.

2. Step 2: Verify container types and loops

   Using index loop and unordered_map is valid and works as intended.

3. Final Answer:

   No error; code works correctly -> Option C

4. Quick Check:

   Increment freq with freq[key]++ works [OK]

Hint: freq[key]++ increments count safely [OK]

Common Mistakes:

• Thinking freq[key] = freq[key] + 1 causes error
• Confusing unordered_map with unordered_set
• Believing range-based for is mandatory

5. Given two strings s1 and s2, which approach using frequency counters correctly checks if s2 is an anagram of s1?

std::unordered_map freq1, freq2;
for (char c : s1) freq1[c]++;
for (char c : s2) freq2[c]++;
// What next?

hard

A. Compare freq1 and freq2 maps for equality

B. Check if freq1 contains all keys of freq2 only

C. Check if freq2 has more keys than freq1

D. Sort s1 and s2 and compare strings

Solution

1. Step 1: Build frequency maps for both strings

   freq1 and freq2 store counts of each character in s1 and s2 respectively.

2. Step 2: Compare frequency maps for exact match

   If freq1 and freq2 are equal, s2 is an anagram of s1 because they have same chars with same counts.

3. Final Answer:

   Compare freq1 and freq2 maps for equality -> Option A

4. Quick Check:

   Anagram check = freq maps equal [OK]

Hint: Anagrams have identical frequency maps [OK]

Common Mistakes:

• Only checking keys without counts
• Assuming more keys means anagram
• Using sorting instead of frequency maps