Character Frequency Counting in DSA Javascript - Build from Scratch

Character Frequency Counting

📖 Scenario: Imagine you are building a simple tool that counts how many times each letter appears in a word. This can help in games like Scrabble or in analyzing text.

🎯 Goal: You will create a program that counts the frequency of each character in a given word and shows the result as a dictionary-like object.

📋 What You'll Learn

Create a string variable with a specific word

Create an empty object to store character counts

Use a for loop to count each character's frequency

Print the final frequency object

💡 Why This Matters

🌍 Real World

Counting character frequency is useful in text analysis, games, and data compression.

💼 Career

Understanding how to count and store frequencies helps in programming tasks like parsing, searching, and building efficient algorithms.

1

Create the word string

Create a string variable called word and set it to the exact value "balloon".

DSA Javascript

// Your code here

Hint

Use const to create the variable word and assign the string "balloon".

2

Create an empty object for counts

Create an empty object called charCounts to store the frequency of each character.

DSA Javascript

const word = "balloon";
// Your code here

Hint

Use {} to create an empty object and assign it to charCounts.

3

Count characters using a for loop

Use a for loop with the variable char to go through each character in word. Inside the loop, increase the count of char in charCounts by 1, or set it to 1 if it does not exist yet.

DSA Javascript

const word = "balloon";
const charCounts = {};
// Your code here

Hint

Use for (const char of word) to loop through each character. Use an if to check if charCounts[char] exists.

4

Print the character counts

Print the charCounts object using console.log(charCounts) to show the frequency of each character.

DSA Javascript

const word = "balloon";
const charCounts = {};
for (const char of word) {
    if (charCounts[char]) {
        charCounts[char] += 1;
    } else {
        charCounts[char] = 1;
    }
}
// Your code here

Hint

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

Practice

1. What does character frequency counting mean in simple terms?

easy

A. Sorting characters alphabetically

B. Counting how many times each character appears in a string

C. Replacing characters with numbers

D. Removing duplicate characters from a string

Solution

1. Step 1: Understand the meaning of frequency

   Frequency means how often something appears.

2. Step 2: Apply to characters in a string

   Counting frequency means counting how many times each character shows up in the string.

3. Final Answer:

   Counting how many times each character appears in a string -> Option B

4. Quick Check:

   Frequency counting = Counting characters [OK]

Hint: Frequency means count occurrences of each character [OK]

Common Mistakes:

• Confusing frequency with sorting
• Thinking frequency means removing duplicates
• Mixing frequency with character replacement

2. Which of the following JavaScript code snippets correctly initializes an empty object to store character frequencies?

easy

A. const freq = '';

B. const freq = [];

C. const freq = {};

D. const freq = null;

Solution

1. Step 1: Identify the data structure for frequency counting

   We use an object to store characters as keys and counts as values.

2. Step 2: Check correct syntax for empty object

   Empty object is created with curly braces {} in JavaScript.

3. Final Answer:

   const freq = {}; -> Option C

4. Quick Check:

   Empty object = {} [OK]

Hint: Use curly braces {} to create empty object [OK]

Common Mistakes:

• Using array [] instead of object {}
• Using empty string '' instead of object
• Assigning null which is not an object

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

const str = 'hello';
const freq = {};
for (const ch of str) {
  freq[ch] = (freq[ch] || 0) + 1;
}
console.log(freq);

medium

A. {"h":1,"e":1,"l":2,"o":1}

B. {"h":1,"e":1,"l":1,"o":1}

C. {"h":0,"e":0,"l":0,"o":0}

D. SyntaxError

Solution

1. Step 1: Loop through each character in 'hello'

   Characters are 'h', 'e', 'l', 'l', 'o'.

2. Step 2: Count frequency using freq[ch] = (freq[ch] || 0) + 1

   For each character, add 1 to its count or start at 1 if not present.

3. Final Answer:

   {"h":1,"e":1,"l":2,"o":1} -> Option A

4. Quick Check:

   Count 'l' twice, others once [OK]

Hint: Use freq[ch] = (freq[ch] || 0) + 1 to count [OK]

Common Mistakes:

• Counting 'l' only once
• Getting syntax error from code
• Confusing keys and values in output

4. Find the error in this code that counts character frequency:

const str = 'abc';
const freq = {};
for (let i = 0; i <= str.length; i++) {
  const ch = str[i];
  freq[ch] = (freq[ch] || 0) + 1;
}
console.log(freq);

medium

A. Loop condition should be i < str.length, not i <= str.length

B. freq should be initialized as an array, not object

C. Use let instead of const for freq

D. Missing semicolon after freq[ch] assignment

Solution

1. Step 1: Check loop range

   Loop runs from i = 0 to i <= str.length, which means i goes up to str.length (3 for 'abc').

2. Step 2: Identify out-of-bounds access

   str[3] is undefined, so freq[undefined] is updated causing wrong key.

3. Final Answer:

   Loop condition should be i < str.length, not i <= str.length -> Option A

4. Quick Check:

   Array index max is length - 1 [OK]

Hint: Use i < length to avoid out-of-bounds errors [OK]

Common Mistakes:

• Using <= instead of < in loop
• Initializing freq as array instead of object
• Ignoring undefined character access

5. Given the string "aabbccddeeffgghhii", which code snippet correctly counts only characters that appear more than once and stores them in freq object?

hard

A. const freq = {};
for (const ch of str) {
freq[ch] = (freq[ch] || 0) + 1;
}
const filtered = {};
for (const ch in freq) {
if (freq[ch] > 1) filtered[ch] = freq[ch];
}

B. const freq = {};
for (const ch of str) {
if (freq[ch]) freq[ch]++;
else freq[ch] = 1;
}
for (const ch in freq) {
if (freq[ch] < 2) freq[ch] = 0;
}

C. const freq = {};
for (const ch of str) {
freq[ch] = 1;
}
for (const ch in freq) {
if (freq[ch] > 1) delete freq[ch];
}

D. const freq = {};
for (const ch of str) {
freq[ch] = (freq[ch] || 0) + 1;
}
for (const ch in freq) {
if (freq[ch] <= 1) delete freq[ch];
}

Solution

1. Step 1: Count all character frequencies

   Use freq[ch] = (freq[ch] || 0) + 1 to count all characters.

2. Step 2: Delete characters with frequency <= 1 from freq

   Iterate and delete keys where freq[ch] <= 1 to leave only duplicates in freq.

3. Final Answer:

   Code snippet that counts all then deletes singles from freq -> Option D

4. Quick Check:

   Delete freq <= 1 from freq [OK]

Hint: Count all, then delete characters with count <= 1 [OK]

Common Mistakes:

• Deleting keys while iterating freq object
• Setting frequency to 0 instead of removing
• Overwriting counts incorrectly