Anagram Check Techniques in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is an anagram?

An anagram is a word or phrase formed by rearranging the letters of another word or phrase, using all the original letters exactly once.

beginner

How can sorting be used to check if two strings are anagrams?

By sorting both strings alphabetically and comparing them. If the sorted strings are identical, they are anagrams.

intermediate

Explain the frequency count method for anagram checking.

Count the frequency of each character in both strings using a map or array. If all character counts match, the strings are anagrams.

intermediate

Why is the frequency count method often more efficient than sorting for anagram checks?

Because counting frequencies takes linear time O(n), while sorting takes O(n log n), making frequency count faster for large strings.

beginner

What is a quick way to rule out two strings as not anagrams before detailed checks?

Check if their lengths are different. If lengths differ, they cannot be anagrams.

Which of these is NOT a valid technique to check if two strings are anagrams?

A Checking if one string is a substring of the other

B Counting character frequencies and comparing

C Sorting both strings and comparing

D Comparing string lengths first

Checking if one string is a substring of the other does not confirm anagrams.

What is the time complexity of sorting-based anagram check for strings of length n?

A O(n log n)

B O(log n)

C O(n)

D O(n^2)

Sorting takes O(n log n) time for strings of length n.

In frequency count method, what data structure is commonly used to store character counts?

A Stack

B Queue

C Hash map or array

D Linked list

Hash maps or arrays efficiently store character frequencies.

If two strings have different lengths, what can we conclude?

A They might be anagrams

B They cannot be anagrams

C They are definitely anagrams

D Length does not matter

Anagrams must have the same length.

Which method is generally faster for very long strings when checking anagrams?

A Checking substrings

B Sorting both strings

C Reversing one string

D Frequency count of characters

Frequency count runs in linear time, faster than sorting for long strings.

Describe two common techniques to check if two strings are anagrams and explain their differences.

Think about sorting vs counting characters.

Explain why checking string length is a useful first step in anagram checking.

Length mismatch means no anagram.

Practice

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

easy

A. They have the same length but different characters

B. They start with the same letter

C. They contain the same characters in any order

D. They have no characters in common

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 C

4. Quick Check:

   Anagrams = same characters [OK]

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

Common Mistakes:

• Thinking same length means anagrams
• Assuming same first letter means anagrams
• Ignoring character counts

2. Which of the following JavaScript code snippets correctly checks if two strings are anagrams by sorting?

easy

A. const areAnagrams = (a, b) => a.split('').sort() === b.split('').sort();

B. const areAnagrams = (a, b) => a.split().sort().join() === b.split().sort().join();

C. const areAnagrams = (a, b) => a.sort() === b.sort();

D. const areAnagrams = (a, b) => a.split('').sort().join('') === b.split('').sort().join('');

Solution

1. Step 1: Understand string and array methods

   Strings need to be split into arrays to sort, then joined back to strings for comparison.

2. Step 2: Check each option's correctness

   const areAnagrams = (a, b) => a.split('').sort().join('') === b.split('').sort().join(''); correctly splits, sorts, and joins both strings before comparing. const areAnagrams = (a, b) => a.split('').sort() === b.split('').sort(); compares arrays directly (wrong). const areAnagrams = (a, b) => a.sort() === b.sort(); tries to sort strings directly (invalid). const areAnagrams = (a, b) => a.split().sort().join() === b.split().sort().join(); uses split() without argument and join() without argument, which is incorrect.

3. Final Answer:

   const areAnagrams = (a, b) => a.split('').sort().join('') === b.split('').sort().join(''); -> Option D

4. Quick Check:

   Split + sort + join = correct syntax [OK]

Hint: Remember to join sorted arrays before comparing strings [OK]

Common Mistakes:

• Comparing arrays directly with ===
• Not joining arrays back to strings
• Using split() without arguments

3. What will be the output of the following code?

function isAnagram(s1, s2) {
  if (s1.length !== s2.length) return false;
  const count = {};
  for (const char of s1) {
    count[char] = (count[char] || 0) + 1;
  }
  for (const char of s2) {
    if (!count[char]) return false;
    count[char]--;
  }
  return true;
}
console.log(isAnagram('listen', 'silent'));
console.log(isAnagram('hello', 'bello'));

medium

A. true false

B. false true

C. true true

D. false false

Solution

1. Step 1: Analyze the first call isAnagram('listen', 'silent')

   Both strings have same length. Counting characters in 'listen' and subtracting counts for 'silent' matches perfectly, so returns true.

2. Step 2: Analyze the second call isAnagram('hello', 'bello')

   Strings have same length but 'hello' has 'h' which 'bello' does not. The count check fails and returns false.

3. Final Answer:

   true false -> Option A

4. Quick Check:

   Matching counts = true, mismatch = false [OK]

Hint: Count chars in first, subtract in second, check zero [OK]

Common Mistakes:

• Ignoring length check
• Not decrementing counts properly
• Returning true too early

4. The following code is intended to check if two strings are anagrams. What is the bug?

function checkAnagram(a, b) {
  const map = new Map();
  for (let i = 0; i < a.length; i++) {
    map.set(a[i], (map.get(a[i]) || 0) + 1);
    map.set(b[i], (map.get(b[i]) || 0) - 1);
  }
  for (const val of map.values()) {
    if (val !== 0) return false;
  }
  return true;
}

medium

A. It incorrectly updates counts for both strings in the same loop

B. It does not check string lengths before processing

C. It uses Map instead of a plain object which is invalid

D. It forgets to return true at the end

Solution

1. Step 1: Identify missing length check

   The code lacks an initial check to ensure both strings have the same length.

2. Step 2: Understand the impact of missing length check

   If `a.length < b.length`, the loop only iterates `a.length` times, incrementing counts for all of `a` but decrementing only the first `a.length` characters of `b`. Extra characters in `b` are ignored, potentially returning `true` incorrectly. Example: `checkAnagram('ab', 'abc')` balances 'a' and 'b' but ignores 'c', wrongly returns `true`. If `a` longer, it may detect mismatch via extra increments or `undefined` key.

3. Final Answer:

   It does not check string lengths before processing -> Option B

4. Quick Check:

   Missing length check = false positives [OK]

Hint: Check lengths first, then count characters [OK]

Common Mistakes:

• Updating counts for both strings simultaneously
• Assuming Map is invalid in JS
• Missing length check

5. You want to check if two strings are anagrams ignoring spaces and case. Which approach correctly implements this in JavaScript?

function isAnagramAdvanced(s1, s2) {
  // Fill in the blanks
}

hard

A. Normalize strings by removing spaces and lowercasing, then compare sorted characters

B. Compare lengths directly after trimming spaces only

C. Use a Map to count characters but ignore case and spaces only in one string

D. Check if s1 includes s2 ignoring case and spaces

Solution

1. Step 1: Normalize both strings

   Remove all spaces and convert both strings to lowercase to ignore case and spaces.

2. Step 2: Sort and compare normalized strings

   Sort the characters of both normalized strings and compare them to check if they are anagrams.

3. Final Answer:

   Normalize strings by removing spaces and lowercasing, then compare sorted characters -> Option A

4. Quick Check:

   Normalize + sort + compare = correct approach [OK]

Hint: Clean strings first, then sort and compare [OK]

Common Mistakes:

• Ignoring spaces in one string only
• Comparing lengths without normalization
• Using includes() which is incorrect for anagrams