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DSA C#programming~5 mins

Anagram Check Techniques in DSA C# - Cheat Sheet & Quick Revision

Choose your learning style10 modes available
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.
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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.
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intermediate
Explain the frequency counting method for anagram checking.
Count the frequency of each character in both strings using a dictionary or array. If the counts match for all characters, the strings are anagrams.
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intermediate
Why is the frequency counting method often more efficient than sorting for anagram checks?
Frequency counting runs in linear time O(n), while sorting takes O(n log n). So frequency counting is faster for large strings.
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intermediate
What edge cases should you consider when checking for anagrams?
Consider case sensitivity, spaces, punctuation, and different string lengths. Usually, ignore spaces and case for meaningful anagram checks.
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Which of these is a valid method to check if two strings are anagrams?
ACheck if one string contains the other
BCheck if strings have the same first letter
CCompare string lengths only
DSort both strings and compare
What is the time complexity of the frequency counting method for anagram checking?
AO(n log n)
BO(n)
CO(n^2)
DO(1)
If two strings have different lengths, can they be anagrams?
ANo, never
BOnly if one is a substring of the other
CYes, always
DOnly if they have the same first letter
Which data structure is commonly used for frequency counting in anagram checks?
AStack
BQueue
CDictionary or array
DLinked list
What should you do with spaces and capitalization when checking for anagrams in phrases?
AIgnore spaces and case
BConsider spaces but ignore case
CConsider case but ignore spaces
DConsider both spaces and case
Describe two common techniques to check if two strings are anagrams.
Think about rearranging letters or counting how many times each letter appears.
You got /2 concepts.
    What are important edge cases to consider when implementing an anagram check?
    Think about what might make two strings look different but still be anagrams.
    You got /3 concepts.

      Practice

      (1/5)
      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
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