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

Frequency Counter Pattern Using Hash Map

📖 Scenario: You are working in a small bookstore. You want to count how many times each book title appears in a list of sold books to understand which books are popular.

🎯 Goal: Build a frequency counter using a hash map (JavaScript object) to count how many times each book title appears in the sales list.

📋 What You'll Learn

Create an array called soldBooks with exact book titles given

Create an empty object called bookFrequency to store counts

Use a for loop with variable book to iterate over soldBooks

Inside the loop, update bookFrequency to count each book's occurrences

Print the bookFrequency object at the end

💡 Why This Matters

🌍 Real World

Counting how many times items appear in a list is useful in sales, surveys, and data analysis.

💼 Career

Frequency counters are common in programming jobs for data processing, analytics, and building recommendation systems.

1

Create the sales data array

Create an array called soldBooks with these exact book titles in order: 'The Hobbit', '1984', 'The Hobbit', 'The Catcher in the Rye', '1984', 'The Hobbit'.

DSA Javascript

// Create the soldBooks array with the exact titles
// Your code here

Hint

Use square brackets [] to create an array and include the exact book titles as strings separated by commas.

2

Create the frequency counter object

Create an empty object called bookFrequency to store the count of each book.

DSA Javascript

const soldBooks = ['The Hobbit', '1984', 'The Hobbit', 'The Catcher in the Rye', '1984', 'The Hobbit'];

// Create an empty object called bookFrequency
// Your code here

Hint

Use curly braces {} to create an empty object and assign it to bookFrequency.

3

Count the frequency of each book

Use a for loop with variable book to iterate over soldBooks. Inside the loop, update bookFrequency[book] to count how many times each book appears. If the book is not yet in bookFrequency, set its count to 1; otherwise, add 1 to its existing count.

DSA Javascript

const soldBooks = ['The Hobbit', '1984', 'The Hobbit', 'The Catcher in the Rye', '1984', 'The Hobbit'];
const bookFrequency = {};

// Use a for loop to count each book's frequency
// Your code here

Hint

Use for (const book of soldBooks) to loop. Check if bookFrequency[book] exists. If yes, add 1; if no, set to 1.

4

Print the frequency counter

Print the bookFrequency object using console.log(bookFrequency) to see the count of each book.

DSA Javascript

const soldBooks = ['The Hobbit', '1984', 'The Hobbit', 'The Catcher in the Rye', '1984', 'The Hobbit'];
const bookFrequency = {};

for (const book of soldBooks) {
    if (bookFrequency[book]) {
        bookFrequency[book] += 1;
    } else {
        bookFrequency[book] = 1;
    }
}

// Print the bookFrequency object
// Your code here

Hint

Use console.log(bookFrequency) to print the object showing counts.

Practice

1. What is the main purpose of using a frequency counter pattern with a hash map in JavaScript?

easy

A. To reverse the order of elements in an array

B. To sort the elements in ascending order

C. To remove duplicate items from a list

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

Solution

1. Step 1: Understand frequency counter pattern

   The frequency counter pattern uses a hash map to store counts of each item in a collection.

2. Step 2: Identify the purpose

   Counting occurrences helps to quickly check how many times each item appears.

3. Final Answer:

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

4. Quick Check:

   Frequency counter = count items [OK]

Hint: Frequency counter counts items using a hash map [OK]

Common Mistakes:

• Confusing frequency counting with sorting
• Thinking it removes duplicates instead of counting
• Assuming it reverses arrays

2. Which of the following is the correct way to initialize an empty frequency counter object in JavaScript?

easy

A. const freq = [];

B. const freq = {};

C. const freq = null;

D. const freq = 0;

Solution

1. Step 1: Identify correct data structure for frequency counter

   A frequency counter uses an object (hash map) to store keys and counts.

2. Step 2: Choose correct initialization syntax

   In JavaScript, an empty object is created with curly braces {}.

3. Final Answer:

   const freq = {}; -> Option B

4. Quick Check:

   Empty object = {} [OK]

Hint: Use {} to create empty object for frequency counter [OK]

Common Mistakes:

• Using array [] instead of object {}
• Assigning null or 0 which are not objects
• Confusing object with other data types

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

const arr = ['a', 'b', 'a', 'c', 'b', 'a'];
const freq = {};
for (const char of arr) {
  freq[char] = (freq[char] || 0) + 1;
}
console.log(freq);

medium

A. {"a": 3, "b": 2, "c": 1}

B. {"a": 1, "b": 1, "c": 1}

C. {"a": 2, "b": 3, "c": 1}

D. {"a": 3, "b": 3, "c": 1}

Solution

1. Step 1: Loop through array and count characters

   Each character's count is incremented in freq object: 'a' appears 3 times, 'b' 2 times, 'c' 1 time.

2. Step 2: Construct final frequency object

   freq = {"a": 3, "b": 2, "c": 1} after loop ends.

3. Final Answer:

   {"a": 3, "b": 2, "c": 1} -> Option A

4. Quick Check:

   Count 'a' = 3, 'b' = 2, 'c' = 1 [OK]

Hint: Count each char incrementally using freq[char] = (freq[char]||0)+1 [OK]

Common Mistakes:

• Mixing counts of 'a' and 'b'
• Not initializing counts properly
• Using arrays instead of objects for counting

4. Identify the error in the following frequency counter code and select the fix:

const arr = ['x', 'y', 'x', 'z'];
const freq = {};
for (let i = 0; i < arr.length; i++) {
  freq[arr[i]] += 1;
}
console.log(freq);

medium

A. Initialize freq[arr[i]] to 0 before adding 1

B. Change for loop to for...of loop

C. Use freq.push(arr[i]) instead of freq[arr[i]]

D. Declare freq as an array instead of object

Solution

1. Step 1: Understand the error cause

   freq[arr[i]] is undefined initially, so adding 1 causes NaN.

2. Step 2: Fix by initializing count

   Set freq[arr[i]] = 0 if undefined before adding 1.

3. Final Answer:

   Initialize freq[arr[i]] to 0 before adding 1 -> Option A

4. Quick Check:

   Initialize before increment to avoid NaN [OK]

Hint: Always initialize count to 0 before incrementing [OK]

Common Mistakes:

• Trying to push to object as if array
• Ignoring undefined initial values
• Changing loop type unnecessarily

5. Given two arrays, arr1 = [1, 2, 2, 3] and arr2 = [2, 3, 3, 4], which code snippet correctly checks if arr2 contains the same frequency of elements as arr1 using the frequency counter pattern?

hard

A. const freq1 = {};
const freq2 = {};
for (const num of arr1) freq1[num] = (freq1[num] || 0) + 1;
for (const num of arr2) freq2[num] = (freq2[num] || 0) + 1;
for (const key in freq1) {
if (freq1[key] !== freq2[key]) return false;
}
return true;

B. const freq = {};
for (const num of arr1) freq[num] = 1;
for (const num of arr2) freq[num] = 0;
return Object.values(freq).every(v => v === 0);

C. const freq = {};
for (const num of arr1) freq[num] = (freq[num] || 0) + 1;
for (const num of arr2) freq[num] -= 1;
for (const key in freq) {
if (freq[key] !== 0) return false;
}
return true;

D. const freq1 = new Map();
const freq2 = new Map();
arr1.forEach(num => freq1.set(num, (freq1.get(num) || 0) + 1));
arr2.forEach(num => freq2.set(num, (freq2.get(num) || 0) + 1));
return [...freq1.entries()].every(([key, val]) => freq2.get(key) === val);

Solution

1. Step 1: Understand frequency comparison logic

   We must count frequencies in both arrays and compare counts for each element.

2. Step 2: Analyze each code snippet

   const freq = {}; for (const num of arr1) freq[num] = (freq[num] || 0) + 1; for (const num of arr2) freq[num] -= 1; for (const key in freq) { if (freq[key] !== 0) return false; } return true; This uses a single object: increments for arr1, decrements for arr2 (NaN for extras), checks all zero. Handles extras (NaN≠0), shortages (positive≠0), overcounts (negative≠0).

3. Final Answer:

   const freq = {}; increment arr1, decrement arr2, check all zero -> Option C

4. Quick Check:

   Single freq: +arr1, -arr2, all zero? [OK]

Hint: Increment arr1 counts, decrement arr2, check all zero for same frequency [OK]

Common Mistakes:

• Not comparing all keys in both frequency counters
• Overwriting counts instead of incrementing
• Using single frequency object incorrectly