Anagram Check Techniques in DSA C# - Time & Space Complexity

Time Complexity: Anagram Check Techniques

O(n log n)

Understanding Time Complexity

We want to understand how the time needed to check if two words are anagrams changes as the words get longer.

How does the work grow when the input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Check if two strings are anagrams by sorting
bool AreAnagrams(string s1, string s2) {
    if (s1.Length != s2.Length) return false;
    var arr1 = s1.ToCharArray();
    var arr2 = s2.ToCharArray();
    Array.Sort(arr1);
    Array.Sort(arr2);
    for (int i = 0; i < arr1.Length; i++) {
        if (arr1[i] != arr2[i]) return false;
    }
    return true;
}
    

This code checks if two strings are anagrams by sorting their characters and comparing them.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting both character arrays and then comparing them.
• How many times: Sorting runs in about n log n time, and the comparison loop runs n times, where n is the string length.

How Execution Grows With Input

As the strings get longer, sorting takes more time, growing faster than just checking each character.

Input Size (n)	Approx. Operations
10	About 10 * log(10) ≈ 33 operations
100	About 100 * log(100) ≈ 664 operations
1000	About 1000 * log(1000) ≈ 9965 operations

Pattern observation: The work grows faster than the input size but slower than the square of the input size.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a bit faster than the length of the strings but not as fast as checking every pair of characters.

Common Mistake

[X] Wrong: "Comparing characters one by one without sorting is always faster."

[OK] Correct: Without sorting or counting, you might miss differences in character counts, so a simple one-by-one check can fail to detect anagrams correctly.

Interview Connect

Understanding how sorting affects time helps you explain your choices clearly and shows you know how to balance speed and correctness.

Self-Check

"What if we used a character count array instead of sorting? How would the time complexity change?"

Practice

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

easy

A. They have the same characters in the same frequency but possibly different order

B. They have the same length but different characters

C. They start with the same letter

D. They contain only vowels

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 with the same counts.

2. Step 2: Compare the options

   Only They have the same characters in the same frequency but possibly different order correctly describes this property; others describe unrelated conditions.

3. Final Answer:

   They have the same characters in the same frequency but possibly different order -> Option A

4. Quick Check:

   Anagram = Same chars & frequency [OK]

Hint: Anagrams have same letters and counts, order doesn't matter [OK]

Common Mistakes:

• Thinking same length means anagram
• Assuming order must be identical
• Confusing anagrams with substrings

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

easy

A. return s1.Contains(s2);

B. return s1 == s2;

C. return s1.Length == s2.Length;

D. return string.Concat(s1.OrderBy(c => c)) == string.Concat(s2.OrderBy(c => c));

Solution

1. Step 1: Understand sorting approach for anagrams

   Sorting both strings and comparing them checks if they have the same characters in the same order after sorting.

2. Step 2: Analyze each code snippet

   return string.Concat(s1.OrderBy(c => c)) == string.Concat(s2.OrderBy(c => c)); sorts both strings and compares; this is correct. Options B, C, and D do not check anagram conditions properly.

3. Final Answer:

   return string.Concat(s1.OrderBy(c => c)) == string.Concat(s2.OrderBy(c => c)); -> Option D

4. Quick Check:

   Sort & compare strings = correct anagram check [OK]

Hint: Sort both strings and compare to check anagrams [OK]

Common Mistakes:

• Comparing strings directly without sorting
• Only checking length equality
• Using Contains instead of full comparison

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

string s1 = "listen";
string s2 = "silent";
bool result = s1.OrderBy(c => c).SequenceEqual(s2.OrderBy(c => c));
Console.WriteLine(result);

medium

A. True

B. False

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Sort both strings and compare sequences

   The code sorts characters of both strings and checks if sequences are equal. "listen" and "silent" are anagrams.

2. Step 2: Evaluate the result of SequenceEqual

   Since sorted characters match, SequenceEqual returns true, so Console.WriteLine prints True.

3. Final Answer:

   True -> Option A

4. Quick Check:

   Sorted anagrams equal = True [OK]

Hint: Sorted anagrams produce true when compared with SequenceEqual [OK]

Common Mistakes:

• Assuming order matters before sorting
• Confusing SequenceEqual with Equals
• Expecting false because strings differ initially

4. Identify the bug in this C# method intended to check anagrams:

bool AreAnagrams(string s1, string s2) {
  if (s1.Length != s2.Length) return false;
  var dict = new Dictionary();
  foreach (var c in s1) {
    dict[c] = dict.ContainsKey(c) ? dict[c] + 1 : 1;
  }
  foreach (var c in s2) {
    if (!dict.ContainsKey(c)) return false;
    dict[c]--;
  }
  return true;
}

medium

A. It does not check string lengths

B. It uses a dictionary instead of array

C. It does not check if counts go below zero

D. It returns false too early

Solution

1. Step 1: Review dictionary count decrement logic

   The method decrements counts for characters in s2 but never checks if count becomes negative, which means extra characters in s2 are not detected.

2. Step 2: Identify missing validation

   Without checking if dict[c] < 0 after decrement, method can wrongly return true when s2 has extra characters.

3. Final Answer:

   It does not check if counts go below zero -> Option C

4. Quick Check:

   Check negative counts after decrement to fix bug [OK]

Hint: Always verify counts don't go negative when decrementing [OK]

Common Mistakes:

• Ignoring negative counts after decrement
• Assuming length check is enough
• Not using dictionary for frequency

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

hard

A. Count characters without removing spaces

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

C. Compare strings directly ignoring case

D. Check if lengths are equal ignoring spaces

Solution

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

   This ensures spaces and case differences do not affect comparison.

2. Step 2: Sort normalized strings and compare

   Sorting both normalized strings and comparing confirms if they are anagrams ignoring case and spaces.

3. Final Answer:

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

4. Quick Check:

   Normalize then sort & compare for case/space insensitive anagram [OK]

Hint: Normalize strings by removing spaces and lowercasing before checking [OK]

Common Mistakes:

• Not removing spaces before comparison
• Ignoring case differences
• Only checking length equality