Frequency Counter Pattern Using Hash Map in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Frequency Counter for Characters

What is the output of the following JavaScript code that counts character frequencies in a string?

DSA Javascript

const str = "banana";
const freq = {};
for (const char of str) {
  freq[char] = (freq[char] || 0) + 1;
}
console.log(freq);

A {"b":1,"a":3,"n":2}

B {"b":1,"a":2,"n":3}

C {"b":1,"a":3,"n":3}

D {"b":1,"a":2,"n":2}

💡 Hint

Count how many times each letter appears in "banana".

The string "banana" has 1 'b', 3 'a's, and 2 'n's. The code counts each character's occurrences correctly.

❓ Predict Output

intermediate

Frequency Counter for Array Elements

What will be printed after running this JavaScript code that counts frequencies of numbers in an array?

DSA Javascript

const arr = [2, 3, 2, 4, 3, 2];
const freq = {};
for (const num of arr) {
  freq[num] = (freq[num] || 0) + 1;
}
console.log(freq);

A {"2":1,"3":2,"4":3}

B {"2":2,"3":2,"4":2}

C {"2":3,"3":3,"4":1}

D {"2":3,"3":2,"4":1}

💡 Hint

Count how many times each number appears in the array.

Number 2 appears 3 times, 3 appears 2 times, and 4 appears once.

🧠 Conceptual

advanced

Frequency Counter Pattern Use Case

Which of the following problems is best solved using the frequency counter pattern with a hash map?

A Checking if two strings are anagrams of each other

B Reversing a linked list

C Finding the maximum value in a sorted array

D Calculating the factorial of a number

💡 Hint

Think about comparing character counts between two strings.

Anagrams have the same frequency of each character, so counting frequencies helps verify this efficiently.

🔧 Debug

advanced

Identify the Bug in Frequency Counter Code

What error or issue will occur when running this code snippet?

DSA Javascript

const arr = [1, 2, 2, 3];
const freq = new Map();
for (const num of arr) {
  if (freq.has(num)) {
    freq.set(num, freq.get(num) + 1);
  } else {
    freq[num] = 1;
  }
}
console.log(freq);

A The code will print Map(4) with correct counts

B TypeError because freq.get is not a function

C freq[num] = 1 does not add to Map, so counts are incorrect

D SyntaxError due to incorrect Map usage

💡 Hint

Check how Map objects store values and how to add new keys.

Using freq[num] = 1 adds a property to the object but does not update the Map. The Map remains missing that key.

❓ Predict Output

expert

Frequency Counter with Nested Objects Output

What is the output of this JavaScript code that counts frequencies of words and their first letters?

DSA Javascript

const words = ["apple", "apricot", "banana", "blueberry", "avocado"];
const freq = {};
for (const word of words) {
  const first = word[0];
  freq[first] = freq[first] || {};
  freq[first][word] = (freq[first][word] || 0) + 1;
}
console.log(freq);

A {"a":{"apple":1,"avocado":1},"b":{"banana":1,"blueberry":1,"apricot":1}}

B {"a":{"apple":1,"apricot":1,"avocado":1},"b":{"banana":1,"blueberry":1}}

C {"a":{"apple":1,"apricot":1,"avocado":1,"banana":1},"b":{"blueberry":1}}

D {"a":{"apple":1,"apricot":1},"b":{"banana":1,"blueberry":1,"avocado":1}}

💡 Hint

Group words by their first letter and count each word's frequency.

Words starting with 'a' are apple, apricot, avocado each counted once; words starting with 'b' are banana and blueberry each once.

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