Valid Anagram

1. Clarify the problem:

   • The problem requires determining whether two strings are anagrams of each other.
   • An anagram is a word or phrase formed by rearranging the letters of another word or phrase.
   • We need to write a function that takes in two strings as input and returns true if they are anagrams and false otherwise.

2. Analyze the problem:

   • Input: Two strings, s and t.
   • Output: Boolean value indicating whether s and t are anagrams.
   • Constraints:
     • The input strings only contain lowercase alphabets.
     • The length of the input strings can vary.
     • The comparison should be case-sensitive.

3. Design an algorithm:

   • First, we can check if the lengths of the two strings are equal. If they are not, we can immediately return false.
   • Next, we can create a frequency count of characters for both strings using a dictionary.
   • We can iterate over the characters in the first string and increment the count for each character encountered in the dictionary.
   • Similarly, we can iterate over the characters in the second string and decrement the count for each character encountered.
   • After iterating over both strings, if all the count values in the dictionary are zero, it means the strings have the same characters with the same frequencies, indicating that they are anagrams.
   • If any count value is non-zero, it means the strings have different characters or different frequencies, indicating that they are not anagrams.

4. Explain your approach:

   • To solve the problem, we will use the frequency count approach using a dictionary.
   • We will create an empty dictionary to store the frequency count of characters.
   • We will iterate over both strings to update the frequency count dictionary.
   • Finally, we will check if all the count values in the dictionary are zero to determine if the strings are anagrams.

5. Write clean and readable code:

python

def isAnagram(s, t):
    if len(s) != len(t):
        return False

    frequency = {}

    for char in s:
        frequency[char] = frequency.get(char, 0) + 1

    for char in t:
        frequency[char] = frequency.get(char, 0) - 1

    for count in frequency.values():
        if count != 0:
            return False

    return True

6. Test your code:

   • Test Case 1: isAnagram("anagram", "nagaram")

     • Explanation: Both strings contain the same characters with the same frequencies.
     • Expected Output: True

   • Test Case 2: isAnagram("rat", "car")

     • Explanation: The strings have different characters.
     • Expected Output: False

   • Test Case 3: isAnagram("listen", "silent")

     • Explanation: Both strings contain the same characters with the same frequencies.
     • Expected Output: True

   • Test Case 4: isAnagram("abcdef", "fedcba")

     • Explanation: Both strings contain the same characters with the same frequencies.
     • Expected Output: True

7. Optimize if necessary:

   • The provided solution has a time complexity of O(n), where n is the length of the input strings. This is because we iterate over both strings once to update the frequency count dictionary.
   • The space complexity is O(1) in the best case and O(n) in the worst case, where n is the length of the input strings. This is because the space required for the dictionary depends on the unique characters present in the strings.

8. Handle error cases:

   • The code assumes that the input strings only contain lowercase alphabets. If the input strings contain other characters or uppercase alphabets, the code may not produce the correct result. We can add input validation to handle such cases.

9. Discuss complexity analysis:

   • The time complexity of the solution is O(n) because we iterate over both strings once.
   • The space complexity is O(n) in the worst case, where n is the length of the input strings. This is because the space required for the dictionary depends on the unique characters present in the strings.