Character Frequency Counting in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of character frequency count in a string

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

DSA Javascript

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

A {"h":1,"e":1,"l":3,"o":1," ":1,"w":1,"r":1,"d":1}

B {"h":1,"e":1,"l":2,"o":2," ":1,"w":1,"r":1,"d":1}

C {"h":1,"e":1,"l":3,"o":2," ":1,"w":1,"r":1,"d":1}

D {"h":1,"e":1,"l":3,"o":2,"w":1,"r":1,"d":1}

💡 Hint

Count each character carefully, including spaces.

The code iterates over each character in the string and counts how many times it appears. The letter 'l' appears 3 times and 'o' appears 2 times, including the space character counted once.

❓ Predict Output

intermediate

Character frequency count with case sensitivity

What is the output of this code that counts characters in a case-sensitive way?

DSA Javascript

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

A {"A":2,"a":2,"B":1,"b":1}

B {"A":2,"a":2,"B":2,"b":2}

C {"A":3,"B":2,"a":1,"b":1}

D {"A":2,"a":1,"B":2,"b":1}

💡 Hint

Uppercase and lowercase letters are counted separately.

The code counts 'A' and 'a' as different characters. 'A' appears twice, 'a' appears twice, 'B' once, and 'b' once.

🔧 Debug

advanced

Identify the error in character frequency counting code

What error will this code produce when run?

DSA Javascript

const str = "test";
const freq = new Map();
for (const ch of str) {
  freq[ch] = (freq[ch] || 0) + 1;
}
console.log(freq);

A TypeError: freq[ch] is not a function

B Output: Map(0) {}

C TypeError: freq[ch] is undefined

D Output: Map(4) { 't' => 2, 'e' => 1, 's' => 1 }

💡 Hint

Check how Map objects are accessed and updated.

Map objects do not support bracket notation for setting keys. Using freq[ch] does not update the Map. The Map remains empty.

❓ Predict Output

advanced

Correct character frequency count using Map

What is the output of this code that correctly counts character frequencies using a Map?

DSA Javascript

const str = "banana";
const freq = new Map();
for (const ch of str) {
  freq.set(ch, (freq.get(ch) || 0) + 1);
}
console.log(Object.fromEntries(freq));

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

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

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

D {"b":2,"a":3,"n":2}

💡 Hint

Use Map's get and set methods properly.

The code uses Map methods correctly to count characters. 'b' appears once, 'a' three times, 'n' twice.

🧠 Conceptual

expert

Time complexity of character frequency counting

What is the time complexity of counting character frequencies in a string of length n using a hash map or object in JavaScript?

A O(1) because counting is done instantly

B O(n^2) because each character requires searching the entire string

C O(log n) because hash map operations are logarithmic

D O(n) because each character is processed once and hash map operations are O(1) on average

💡 Hint

Consider how many times each character is visited and the cost of updating counts.

Each character is visited once, and updating the count in a hash map is O(1) on average, so total time is O(n).

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