Anagram Check Techniques in DSA Typescript - Build from Scratch

Anagram Check Techniques

📖 Scenario: You are building a simple tool to check if two words are anagrams. Anagrams are words that have the same letters but in a different order, like "listen" and "silent".This tool will help users quickly find out if two words are anagrams by comparing their letters.

🎯 Goal: Build a TypeScript program that checks if two given strings are anagrams by comparing their sorted characters.

📋 What You'll Learn

Create two string variables with exact values

Create a helper variable to hold sorted characters

Write the core logic to check if the two strings are anagrams

Print the result as true or false

💡 Why This Matters

🌍 Real World

Anagram checking is useful in word games, puzzles, and text analysis tools to find matching words or detect rearrangements.

💼 Career

Understanding string manipulation and comparison is fundamental for software developers working on text processing, search engines, and data validation.

1

Create the two strings to compare

Create two string variables called word1 and word2 with the exact values "listen" and "silent" respectively.

DSA Typescript

// Your code here
// Create word1 and word2 strings

Hint

Use const to create variables and assign the exact strings.

2

Create sorted character arrays for both words

Create two variables called sortedWord1 and sortedWord2 that hold the sorted characters of word1 and word2 respectively. Use split(''), sort(), and join('') methods.

DSA Typescript

const word1: string = "listen";
const word2: string = "silent";
// Your code here
// Create sortedWord1 and sortedWord2 by sorting characters of word1 and word2

Hint

Split the string into characters, sort them, then join back to a string.

3

Check if the two sorted strings are equal

Create a boolean variable called areAnagrams that is true if sortedWord1 equals sortedWord2, otherwise false.

DSA Typescript

const word1: string = "listen";
const word2: string = "silent";
const sortedWord1: string = word1.split('').sort().join('');
const sortedWord2: string = word2.split('').sort().join('');
// Your code here
// Create areAnagrams boolean by comparing sortedWord1 and sortedWord2

Hint

Use the equality operator === to compare the sorted strings.

4

Print the result

Print the value of areAnagrams using console.log.

DSA Typescript

const word1: string = "listen";
const word2: string = "silent";
const sortedWord1: string = word1.split('').sort().join('');
const sortedWord2: string = word2.split('').sort().join('');
const areAnagrams: boolean = sortedWord1 === sortedWord2;
// Your code here
// Print areAnagrams

Hint

Use console.log(areAnagrams); to print the result.

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