Anagram Check Techniques in DSA Java - 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?

Sort both strings alphabetically and compare them. If they are identical after sorting, they are anagrams.

intermediate

Explain the frequency count technique for anagram checking.

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

intermediate

What is the time complexity of the sorting-based anagram check?

The time complexity is O(n log n), where n is the length of the strings, due to sorting.

intermediate

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

Because frequency counting can be done in O(n) time by scanning the strings once, avoiding the overhead of sorting.

Which technique involves sorting both strings to check for anagrams?

A Sorting-based method

B Frequency count method

C Hashing with random keys

D Dynamic programming

Sorting both strings and comparing them is the sorting-based method for anagram checking.

What data structure is commonly used in the frequency count method?

A Stack

B Array or HashMap

C Queue

D Linked List

An array or HashMap is used to count character frequencies in the frequency count method.

If two strings have different lengths, what can we conclude about them being anagrams?

A Length does not matter

B They might be anagrams

C They are definitely anagrams

D They cannot be anagrams

Strings of different lengths cannot be anagrams because anagrams must use all letters exactly once.

What is the time complexity of the frequency count method for anagram checking?

A O(n log n)

B O(1)

C O(n)

D O(n^2)

Frequency count method runs in O(n) time by scanning the strings once.

Which of these is NOT a valid step in checking anagrams using frequency count?

A Sort both strings

B Increment counts for first string

C Decrement counts for second string

D Initialize frequency array

Sorting is not part of the frequency count method; it belongs to the sorting-based method.

Describe two common techniques to check if two strings are anagrams.

Think about rearranging letters and counting letters.

Explain why frequency count method can be more efficient than sorting for anagram checks.

Consider the time complexity of sorting vs counting.

Practice

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

easy

A. They end with the same character

B. They have the same length but different characters

C. They start with the same character

D. They contain the same characters in any order

Solution

1. Step 1: Understand the definition of anagrams

   Anagrams are words or phrases made by rearranging the letters of another, using all original letters exactly once.

2. Step 2: Compare character sets

   If two strings have the same characters with the same frequency, they are anagrams regardless of order.

3. Final Answer:

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

4. Quick Check:

   Anagrams = same characters any order [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 frequency

2. Which Java code snippet correctly checks if two strings s1 and s2 are anagrams using sorting?

easy

A. char[] a1 = s1.toCharArray(); char[] a2 = s2.toCharArray(); Arrays.sort(a1); Arrays.sort(a2); return Arrays.equals(a1, a2);

B. Arrays.sort(s1.toCharArray()); Arrays.sort(s2.toCharArray()); return s1.equals(s2);

C. return s1.equalsIgnoreCase(s2);

D. return s1.length() == s2.length();

Solution

1. Step 1: Convert strings to char arrays and sort

   We must convert strings to char arrays because Arrays.sort works on arrays, not strings.

2. Step 2: Compare sorted arrays for equality

   After sorting, if arrays are equal, strings are anagrams. Use Arrays.equals to compare arrays.

3. Final Answer:

   char[] a1 = s1.toCharArray(); char[] a2 = s2.toCharArray(); Arrays.sort(a1); Arrays.sort(a2); return Arrays.equals(a1, a2); -> Option A

4. Quick Check:

   Sort char arrays and compare arrays [OK]

Hint: Sort char arrays, then compare with Arrays.equals [OK]

Common Mistakes:

• Sorting strings directly without converting to char arrays
• Using equals() on unsorted strings
• Comparing only lengths

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

String s1 = "listen";
String s2 = "silent";
char[] a1 = s1.toCharArray();
char[] a2 = s2.toCharArray();
Arrays.sort(a1);
Arrays.sort(a2);
System.out.println(Arrays.equals(a1, a2));

medium

A. Runtime exception

B. false

C. true

D. Compilation error

Solution

1. Step 1: Convert and sort both strings

   Both strings "listen" and "silent" are converted to char arrays and sorted. Sorted arrays become ['e','i','l','n','s','t'] for both.

2. Step 2: Compare sorted arrays

   Since sorted arrays are identical, Arrays.equals(a1, a2) returns true.

3. Final Answer:

   true -> Option C

4. Quick Check:

   Sorted arrays match = true [OK]

Hint: Sort and compare arrays to check anagrams [OK]

Common Mistakes:

• Assuming order matters without sorting
• Confusing equals() on strings with arrays
• Forgetting to sort arrays

4. Identify the error in this Java method that checks if two strings are anagrams:

public boolean isAnagram(String s1, String s2) {
  if (s1.length() != s2.length()) return false;
  char[] a1 = s1.toCharArray();
  char[] a2 = s2.toCharArray();
  Arrays.sort(a1);
  Arrays.sort(a2);
  return a1.equals(a2);
}

medium

A. Using a1.equals(a2) compares references, not content

B. Missing length check before sorting

C. Sorting arrays is incorrect for anagram check

D. Method should return void, not boolean

Solution

1. Step 1: Understand array equality in Java

   Using a1.equals(a2) compares if both arrays are the same object, not their contents.

2. Step 2: Use correct method to compare arrays

   To compare contents, use Arrays.equals(a1, a2) instead of a1.equals(a2).

3. Final Answer:

   Using a1.equals(a2) compares references, not content -> Option A

4. Quick Check:

   Use Arrays.equals() for array content comparison [OK]

Hint: Use Arrays.equals() to compare array contents [OK]

Common Mistakes:

• Using equals() on arrays instead of Arrays.equals()
• Ignoring length check
• Returning wrong data type

5. You want to check if two strings are anagrams but ignoring case and spaces. Which approach is best?

hard

A. Compare strings directly with equalsIgnoreCase()

B. Remove spaces, convert to lowercase, then sort and compare char arrays

C. Count characters without removing spaces or changing case

D. Check if lengths are equal and first characters match

Solution

1. Step 1: Normalize strings by removing spaces and converting to lowercase

   Removing spaces and converting to lowercase ensures uniform comparison ignoring case and spaces.

2. Step 2: Sort char arrays and compare

   After normalization, convert to char arrays, sort them, and compare to check anagram status.

3. Final Answer:

   Remove spaces, convert to lowercase, then sort and compare char arrays -> Option B

4. Quick Check:

   Normalize then sort and compare [OK]

Hint: Normalize strings before sorting to ignore case and spaces [OK]

Common Mistakes:

• Not removing spaces before comparison
• Ignoring case differences
• Only comparing lengths or first characters