Character Frequency Counting in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize an empty object to store character counts.

DSA Typescript

const charCount = [1];

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

Anull

B[]

C{}

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using an empty array instead of an object.

Initializing with null or zero which cannot hold key-value pairs.

We use an empty object {} to store character counts as key-value pairs.

2fill in blank

medium

Complete the code to loop through each character in the string.

DSA Typescript

for (const [1] of input) {
  // process character
}

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

Aindex

Bi

Cinput

Dchar

Attempts:

3 left

💡 Hint

Common Mistakes

Using index or i which are usually for numeric positions.

Using the whole string variable name which is incorrect.

We use char to represent each character in the string as we loop through it.

3fill in blank

hard

Fix the error in updating the character count in the object.

DSA Typescript

charCount[char] = (charCount[char] || [1]) + 1;

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

A0

BcharCount[char]

C1

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 as default which causes counts to be off by one.

Using null or the character itself which causes errors.

If the character is not yet in the object, we start counting from zero before adding one.

4fill in blank

hard

Fill both blanks to create a frequency map for characters longer than 2.

DSA Typescript

const freqMap = {
  [1]: [2] for (const char of input) if char.length > 2
};

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

Achar

BcharCount[char]

Cchar.length

Dinput

Attempts:

3 left

💡 Hint

Common Mistakes

Using char.length as key or value which is incorrect.

Using input which is the whole string, not a key.

The key is the character itself, and the value is its count from charCount.

5fill in blank

hard

Fill all three blanks to filter and map characters with count greater than 1.

DSA Typescript

const filtered = Object.fromEntries(
  Object.entries(charCount).filter(([[1], [2]]) => [3] > 1)
);

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

Akey

Bvalue

Ccount

Dchar

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent variable names in destructuring.

Comparing the key instead of the value for filtering.

We destructure entries as [char, value], then filter where value (count) is greater than 1.

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