Frequency Counter Pattern Using Hash Map in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create an empty object to count frequencies.

DSA Javascript

const frequency = [1];

Drag options to blanks, or click blank then click option'

A{}

B[]

Cnull

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using [] which creates an array instead of an object.

Using null or 0 which are not objects.

An empty object {} is used to store frequency counts as key-value pairs.

2fill in blank

medium

Complete the code to check if a character exists in the frequency object.

DSA Javascript

if (frequency.hasOwnProperty([1])) {

Drag options to blanks, or click blank then click option'

A'char'

Bchar

Cfrequency

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using a string literal 'char' instead of the variable char.

Using unrelated variables like frequency or i.

Use the variable 'char' to check if that character is a key in the frequency object.

3fill in blank

hard

Fix the error in incrementing the frequency count for a character.

DSA Javascript

frequency[char] [1] 1;

Drag options to blanks, or click blank then click option'

A-=

B++

C=

D+=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' which overwrites the count instead of incrementing.

Using '-=' which decreases the count.

Using '++' incorrectly in this context.

Use '+=' to add 1 to the existing count for the character key.

4fill in blank

hard

Fill both blanks to complete the loop that counts character frequencies.

DSA Javascript

for (let [1] = 0; [2] < str.length; [1]++) {
  const char = str[[1]];
  // count logic here
}

Drag options to blanks, or click blank then click option'

Ai

Bj

Cstr.length

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in initialization and condition.

Using 'length' instead of 'str.length' in the condition.

Use 'i' as the loop variable consistently for initialization and condition.

5fill in blank

hard

Fill all three blanks to create a frequency counter function.

DSA Javascript

function frequencyCounter(str) {
  const frequency = [1];
  for (let i = 0; i < str.length; i++) {
    const char = str[i];
    if (frequency.hasOwnProperty(char)) {
      frequency[char] [2] 1;
    } else {
      frequency[char] = [3];
    }
  }
  return frequency;
}

Drag options to blanks, or click blank then click option'

A{}

B+=

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing frequency as an array or number.

Using '=' instead of '+=' to increment counts.

Setting new keys to 0 instead of 1.

Initialize frequency as an empty object {}, increment counts with '+=', and set new keys to 1.

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