Character Frequency Counting in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of character frequency count in a string

What is the output of the following TypeScript code that counts character frequency in a string?

DSA Typescript

function charFrequency(str: string): Record<string, number> {
  const freq: Record<string, number> = {};
  for (const ch of str) {
    freq[ch] = (freq[ch] || 0) + 1;
  }
  return freq;
}

console.log(charFrequency("hello"));

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

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

C {"h":1,"e":1,"l":3,"o":1}

D {"h":1,"e":2,"l":2,"o":1}

💡 Hint

Count how many times each letter appears in the string "hello".

The function counts each character's occurrences. In "hello", 'l' appears twice, others once.

❓ Predict Output

intermediate

Character frequency with spaces included

What is the output of this TypeScript code counting characters including spaces?

DSA Typescript

function charFrequency(str: string): Record<string, number> {
  const freq: Record<string, number> = {};
  for (const ch of str) {
    freq[ch] = (freq[ch] || 0) + 1;
  }
  return freq;
}

console.log(charFrequency("a b a"));

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

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

C {"a":1," ":2,"b":1}

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

💡 Hint

Remember to count spaces as characters too.

The string "a b a" has two 'a's, one 'b', and two spaces.

🔧 Debug

advanced

Identify the error in character frequency code

What error does this TypeScript code produce when counting character frequency?

DSA Typescript

function charFrequency(str: string): Record<string, number> {
  const freq: Record<string, number> = {};
  for (const ch of str) {
    freq[ch] += 1;
  }
  return freq;
}

console.log(charFrequency("test"));

A TypeScript error: Object is possibly 'undefined'

B Output: {"t":3,"e":1,"s":1}

C Output: {"t":2,"e":1,"s":1}

D SyntaxError: Unexpected token '+'

💡 Hint

Check what happens when freq[ch] is undefined before adding 1.

The code tries to add 1 to freq[ch] without initializing it first, causing a TypeScript compile error because freq[ch] may be undefined.

🚀 Application

advanced

Find the most frequent character

Given the character frequency object, which code snippet correctly finds the character with the highest frequency?

DSA Typescript

const freq = { a: 3, b: 5, c: 2 };
// Find character with max frequency

A

const maxChar = Object.keys(freq).reduce((a, b) => freq[a] < freq[b] ? a : b);
console.log(maxChar);

B

const maxChar = Object.values(freq).reduce((a, b) => a > b ? a : b);
console.log(maxChar);

C

let maxChar = '';
let maxCount = 0;
for (const ch in freq) {
  if (freq[ch] > maxCount) {
    maxCount = freq[ch];
    maxChar = ch;
  }
}
console.log(maxChar);

D

const maxChar = Object.keys(freq).reduce((a, b) => freq[a] > freq[b] ? b : a);
console.log(maxChar);

💡 Hint

Check which option correctly updates maxChar and maxCount to find max frequency.

Option C correctly iterates and updates maxChar when a higher frequency is found.

🧠 Conceptual

expert

Time complexity of character frequency counting

What is the time complexity of counting character frequency in a string of length n using a hash map?

A O(n)

B O(n^2)

C O(log n)

D O(1)

💡 Hint

Consider how many times the loop runs and the cost of each operation.

Counting frequency requires one pass through the string, so it is O(n).

Practice

1. What does character frequency counting in a string help us find?

easy

A. How many times each character appears in the string

B. The length of the string

C. The first character of the string

D. The position of a character in the string

Solution

1. Step 1: Understand character frequency counting

   Character frequency counting means counting how many times each character appears in a string.

2. Step 2: Identify what it helps find

   This counting helps us know the number of occurrences of each character, not the length or position.

3. Final Answer:

   How many times each character appears in the string -> Option A

4. Quick Check:

   Frequency counting = character counts [OK]

Hint: Frequency counting means counting each character's occurrences [OK]

Common Mistakes:

• Confusing frequency with string length
• Thinking it finds character positions
• Mixing frequency with first character only

2. Which TypeScript code snippet correctly initializes an empty object to store character frequencies?

easy

A. const freq = new Map();

B. const freq = {};

C. const freq = [];

D. const freq = '';

Solution

1. Step 1: Identify data structure for frequency counting

   We need an object to store keys (characters) and values (counts). An empty object is initialized with {}.

2. Step 2: Check each option

   const freq = new Map(); uses Map which is valid but not the simplest empty object. const freq = {}; correctly uses {}. const freq = []; is an array, not suitable. const freq = ''; is a string, not suitable.

3. Final Answer:

   const freq = {}; -> Option B

4. Quick Check:

   Empty object = {} [OK]

Hint: Use {} to start an empty object for frequency counts [OK]

Common Mistakes:

• Using array [] instead of object {}
• Using empty string '' instead of object
• Confusing Map with plain object initialization

3. What is the output of this TypeScript code?

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

medium

A. {"h":1,"e":1,"l":3,"o":1}

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

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

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

Solution

1. Step 1: Loop through each character and count

   Characters: 'h', 'e', 'l', 'l', 'o'. For each, increase count in freq object.

2. Step 2: Calculate counts

   'h' = 1, 'e' = 1, 'l' = 2 (appears twice), 'o' = 1.

3. Final Answer:

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

4. Quick Check:

   Count 'l' twice = 2 [OK]

Hint: Add 1 or start at 0 for each char count [OK]

Common Mistakes:

• Counting 'l' only once
• Setting counts to zero
• Miswriting object output format

4. Find the error in this TypeScript code for counting character frequency:

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

medium

A. Use 'for...in' instead of 'for' loop

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

C. Missing declaration of variable ch

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

Solution

1. Step 1: Check loop bounds

   Loop runs from i = 0 to i <= str.length, which means i goes up to str.length (3). But string indices go from 0 to length-1 (2).

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

   At i = 3, str[3] is undefined, causing freq[undefined] to be updated, which is incorrect.

3. Final Answer:

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

4. Quick Check:

   Loop index must be less than length [OK]

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

Common Mistakes:

• Using <= instead of < in loop condition
• Initializing freq as array instead of object
• Confusing for...in with for...of loops

5. Given the string "aabbc", which TypeScript code snippet correctly returns an object with only characters that appear exactly twice?

hard

A. const freq: {[key: string]: number} = {};
for (const ch of str) {
freq[ch] = (freq[ch] || 0) + 1;
}
const result = Object.fromEntries(Object.entries(freq).filter(([k,v]) => v === 2));

B. const freq = new Map();
for (const ch of str) {
freq.set(ch, (freq.get(ch) || 0) + 1);
}
const result = Object.fromEntries([...freq].filter(([k,v]) => v === 1));

C. const freq = {};
for (const ch of str) {
freq[ch] = 2;
}
const result = freq;

D. const freq = {};
for (const ch of str) {
freq[ch] = (freq[ch] || 0) + 1;
}
const result = Object.keys(freq);

Solution

1. Step 1: Count frequencies correctly

   Loop through string, count each character's occurrences in freq object.

2. Step 2: Filter characters with count exactly 2

   Use Object.entries(freq) to get pairs, filter those with value 2, then convert back to object.

3. Final Answer:

   Code snippet that counts and filters for count === 2 -> Option A

4. Quick Check:

   Filter freq entries where value is 2 [OK]

Hint: Filter entries with value === 2 after counting [OK]

Common Mistakes:

• Filtering for count === 1 instead of 2
• Assigning fixed count 2 inside loop
• Returning keys instead of filtered object