Anagram Check Techniques in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if two strings have the same length.

DSA Typescript

function areSameLength(str1: string, str2: string): boolean {
  return str1.length [1] str2.length;
}

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

A===

B!==

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using assignment operator '=' instead of comparison '==='.

Using inequality operators which check for different lengths.

We use === to check if both strings have exactly the same length.

2fill in blank

medium

Complete the code to sort the characters of a string.

DSA Typescript

function sortString(str: string): string {
  return str.split('').[1]().join('');
}

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

Areverse

Bsort

Cmap

Dfilter

Attempts:

3 left

💡 Hint

Common Mistakes

Using reverse() which only reverses order, not sorts.

Using map() or filter() which do not sort.

The sort() method sorts the array of characters alphabetically.

3fill in blank

hard

Fix the error in the code to correctly check if two strings are anagrams by sorting.

DSA Typescript

function areAnagrams(str1: string, str2: string): boolean {
  if (str1.length !== str2.length) return false;
  return str1.split('').[1]().join('') === str2.split('').sort().join('');
}

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

Asort

Breverse

Cfilter

Dmap

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting only one string and not the other.

Using reverse() instead of sort().

Both strings must be sorted to compare their characters correctly.

4fill in blank

hard

Fill both blanks to count characters of a string using a map.

DSA Typescript

function charCount(str: string): Map<string, number> {
  const countMap = new Map<string, number>();
  for (const char of str) {
    countMap.set(char, (countMap.get(char) [1] 0) [2] 1);
  }
  return countMap;
}

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

A||

B+

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Not providing a default value for missing keys.

We use || to default to 0 if the character is not in the map, then add 1.

5fill in blank

hard

Fill all three blanks to check if two strings are anagrams using character counts.

DSA Typescript

function areAnagramsByCount(str1: string, str2: string): boolean {
  if (str1.length !== str2.length) return false;
  const countMap = new Map<string, number>();
  for (const char of str1) {
    countMap.set(char, (countMap.get(char) [1] 0) [2] 1);
  }
  for (const char of str2) {
    if (!countMap.has(char)) return false;
    countMap.set(char, countMap.get(char)! [3] 1);
    if (countMap.get(char)! < 0) return false;
  }
  return true;
}

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

A||

B+

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Not subtracting counts for the second string.

Using multiplication instead of addition or subtraction.

We initialize counts with || 0, add 1 for first string, subtract 1 for second string.

Practice

1. What does it mean if two strings are anagrams of each other?

easy

A. They contain the same characters in any order

B. They have the same length but different characters

C. They start with the same character

D. They end with the same character

Solution

1. Step 1: Understand the definition of anagrams

   Anagrams are words or phrases made by rearranging the letters of another, so they must have the same characters.

2. Step 2: Compare the character sets

   If two strings have exactly the same characters, just in different orders, they are anagrams.

3. Final Answer:

   They contain the same characters in any order -> Option A

4. Quick Check:

   Anagrams = same characters [OK]

Hint: Check if sorted strings match to confirm anagrams [OK]

Common Mistakes:

• Thinking same length means anagrams
• Only comparing first or last characters
• Ignoring character counts

2. Which TypeScript code snippet correctly checks if two strings are anagrams by sorting?

easy

A. const isAnagram = (a: string, b: string) => a.split('').sort().join('') === b.split('').sort().join('');

B. const isAnagram = (a: string, b: string) => a.split('').sort() === b.split('').sort();

C. const isAnagram = (a: string, b: string) => a.sort() === b.sort();

D. const isAnagram = (a: string, b: string) => a === b;

Solution

1. Step 1: Understand string sorting in TypeScript

   Strings must be split into arrays, sorted, then joined back to strings for comparison.

2. Step 2: Identify correct syntax

   const isAnagram = (a: string, b: string) => a.split('').sort().join('') === b.split('').sort().join(''); correctly splits, sorts, and joins both strings before comparing.

3. Final Answer:

   const isAnagram = (a: string, b: string) => a.split('').sort().join('') === b.split('').sort().join(''); -> Option A

4. Quick Check:

   Sort and join strings before compare [OK]

Hint: Always join after sorting arrays to compare strings [OK]

Common Mistakes:

• Comparing arrays directly without join
• Trying to sort strings directly
• Comparing strings without sorting

3. What is the output of this TypeScript code?

function areAnagrams(s1: string, s2: string): boolean {
  const normalize = (str: string) => str.toLowerCase().replace(/\s+/g, '').split('').sort().join('');
  return normalize(s1) === normalize(s2);
}

console.log(areAnagrams('Listen', 'Silent'));
console.log(areAnagrams('Hello', 'Olelh'));
console.log(areAnagrams('Test', 'Taste'));

medium

A. true false false

B. true true false

C. false true true

D. false false true

Solution

1. Step 1: Normalize strings by lowercasing, removing spaces, sorting

   'Listen' and 'Silent' normalize to 'eilnst', 'Hello' and 'Olelh' normalize to 'ehllo', 'Test' and 'Taste' normalize to 'estt' and 'aestt' respectively.

2. Step 2: Compare normalized strings

   'Listen' vs 'Silent' match (true), 'Hello' vs 'Olelh' match (true), 'Test' vs 'Taste' do not match (false).

3. Final Answer:

   true true false -> Option B

4. Quick Check:

   Normalized strings equal means true [OK]

Hint: Normalize strings before comparing for anagram check [OK]

Common Mistakes:

• Ignoring case differences
• Not removing spaces
• Not sorting characters

4. Identify the bug in this TypeScript function that checks anagrams:

function isAnagram(a: string, b: string): boolean {
  return a.split('').sort() === b.split('').sort();
}

medium

A. Not removing spaces before comparison

B. Missing toLowerCase() for case-insensitive check

C. Comparing arrays directly instead of strings

D. Using split() without arguments

Solution

1. Step 1: Understand array comparison in TypeScript

   Arrays compared with === check reference, not content, so two sorted arrays won't be equal even if same elements.

2. Step 2: Fix by joining arrays to strings before comparing

   Joining sorted arrays to strings allows content comparison, fixing the bug.

3. Final Answer:

   Comparing arrays directly instead of strings -> Option C

4. Quick Check:

   Arrays need join() before compare [OK]

Hint: Join sorted arrays before comparing for equality [OK]

Common Mistakes:

• Comparing arrays with ===
• Ignoring case sensitivity
• Not trimming spaces

5. Given two strings, how can you efficiently check if they are anagrams using a character count map in TypeScript? Which code snippet correctly implements this?

hard

A. function isAnagram(a: string, b: string): boolean {
if (a.length !== b.length) return false;
const count: Record = {};
for (const char of a) count[char] = (count[char] || 0) + 1;
for (const char of b) {
if (!count[char]) return false;
count[char]--;
}
return true;
}

B. function isAnagram(a: string, b: string): boolean {
return a.split('').sort().join('') === b.split('').sort().join('');
}

C. function isAnagram(a: string, b: string): boolean {
return a.length === b.length;
}

D. function isAnagram(a: string, b: string): boolean {
const mapA = new Map();
for (const c of a) mapA.set(c, (mapA.get(c) || 0) + 1);
for (const c of b) mapA.set(c, (mapA.get(c) || 0) - 1);
return [...mapA.values()].every(v => v === 0);
}

Solution

1. Step 1: Understand character count map approach

   Count characters in first string, subtract counts using second string, then check all counts zero.

2. Step 2: Analyze each option

   function isAnagram(a: string, b: string): boolean { if (a.length !== b.length) return false; const count: Record = {}; for (const char of a) count[char] = (count[char] || 0) + 1; for (const char of b) { if (!count[char]) return false; count[char]--; } return true; } misses checking if count goes negative; function isAnagram(a: string, b: string): boolean { return a.split('').sort().join('') === b.split('').sort().join(''); } uses sorting (correct but not map); function isAnagram(a: string, b: string): boolean { return a.length === b.length; } only checks length; function isAnagram(a: string, b: string): boolean { const mapA = new Map(); for (const c of a) mapA.set(c, (mapA.get(c) || 0) + 1); for (const c of b) mapA.set(c, (mapA.get(c) || 0) - 1); return [...mapA.values()].every(v => v === 0); } correctly uses Map to count and verify zero counts.

3. Final Answer:

   Option D correctly implements character count map with Map and zero check -> Option D

4. Quick Check:

   Map counts increment/decrement, all zero means anagram [OK]

Hint: Use map to count chars, decrement, then check all zero [OK]

Common Mistakes:

• Not checking for negative counts
• Only comparing lengths
• Using sorting instead of counting