Why Anagram Check Techniques in DSA Typescript?

The Big Idea

Discover how a simple trick can save you from tedious letter comparisons!

The Scenario

Imagine you want to check if two words are made of the same letters, like "listen" and "silent". Doing this by hand means comparing each letter one by one, which is slow and confusing for long words.

The Problem

Manually checking each letter is tiring and easy to make mistakes. You might miss letters or count them wrong, especially if the words are long or have repeated letters.

The Solution

Anagram check techniques use smart ways to compare words quickly, like sorting letters or counting how many times each letter appears. This makes the check fast and accurate, even for long words.

Before vs After

✗ Before

function areAnagramsManual(word1: string, word2: string): boolean {
  // Compare letters one by one manually
  if (word1.length !== word2.length) return false;
  for (let i = 0; i < word1.length; i++) {
    if (!word2.includes(word1[i])) return false;
  }
  return true;
}

✓ After

function areAnagrams(word1: string, word2: string): boolean {
  return word1.split('').sort().join('') === word2.split('').sort().join('');
}

What It Enables

It enables quick and reliable detection of words that are anagrams, which is useful in games, puzzles, and text analysis.

Real Life Example

In word games like Scrabble or crossword puzzles, checking if two words are anagrams helps find valid moves or hints.

Key Takeaways

Manual letter-by-letter checks are slow and error-prone.

Sorting or counting letters makes anagram checks fast and accurate.

Anagram techniques help in games and text processing.

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