The Valid Anagram problem is a classic coding challenge frequently encountered in technical interviews and programming exercises. It involves determining whether two given strings are anagrams of each other. Two strings are considered anagrams if they contain the same characters in the same frequency, but the order of characters may differ. For example, the strings "listen" and "silent" are anagrams because they contain the same characters, each occurring with the same frequency.

In this post, we’ll explore multiple ways to solve the Valid Anagram problem in Python, detailing the various approaches, edge cases, and efficiency considerations.

Problem Definition

Given two strings s and t, write a function isAnagram that determines whether t is an anagram of s.

Example:

Input: s = "anagram", t = "nagaram"
Output: True

Input: s = "rat", t = "car"
Output: False

Constraints:

  • Both s and t consist of lowercase English letters.
  • The length of both strings can be up to (10^5).

Approach 1: Sorting-Based Solution

Explanation:

A straightforward way to solve the problem is to sort both strings and then compare them. If the sorted version of both strings is the same, then the two strings are anagrams. Sorting ensures that characters are arranged in a specific order, making it easy to compare them.

Code Implementation:

def isAnagram(s: str, t: str) -> bool:
    # If the lengths of the strings are not equal, they can't be anagrams
    if len(s) != len(t):
        return False
    
    # Sort both strings and compare
    return sorted(s) == sorted(t)

# Test cases
print(isAnagram("anagram", "nagaram"))  # Output: True
print(isAnagram("rat", "car"))  # Output: False

Time and Space Complexity:

  • Time Complexity: O(n log n), where n is the length of the string. This is due to the sorting operation, which typically takes O(n log n).
  • Space Complexity: O(n), because sorting generates new copies of the strings.

Explanation:

This approach sorts both strings and compares them. If the sorted versions of the strings are identical, then they must be anagrams. While simple, this solution can be slow for very large strings due to the sorting process.

Drawbacks:

The sorting approach is not the most efficient method because sorting has a time complexity of O(n log n). For large inputs, this can become a bottleneck, making the solution less optimal.

Approach 2: Frequency Count Using Hash Maps (Optimal Solution)

Explanation:

A more efficient way to check if two strings are anagrams is to compare the frequency of characters in both strings. We can use a hash map (in Python, a dictionary) to count the occurrences of each character in both strings. If both strings have the same character frequencies, then they are anagrams.

Code Implementation:

from collections import defaultdict

def isAnagram(s: str, t: str) -> bool:
    # If the lengths of the strings are not equal, they can't be anagrams
    if len(s) != len(t):
        return False

    # Create frequency dictionaries for both strings
    char_count_s = defaultdict(int)
    char_count_t = defaultdict(int)

    # Count characters for both strings
    for char in s:
        char_count_s[char] += 1
    for char in t:
        char_count_t[char] += 1

    # Compare frequency dictionaries
    return char_count_s == char_count_t

# Test cases
print(isAnagram("anagram", "nagaram"))  # Output: True
print(isAnagram("rat", "car"))  # Output: False

Time and Space Complexity:

  • Time Complexity: O(n), where n is the length of the strings. This is because we loop through each string exactly once.
  • Space Complexity: O(n), since we use two dictionaries to store the frequency counts of the characters.

Explanation:

This solution uses two dictionaries to count the frequency of each character in both strings. If the dictionaries are identical, the strings are anagrams. This approach is more efficient than the sorting-based approach because it avoids the O(n log n) complexity of sorting, reducing the time complexity to O(n).

Benefits:

This is an optimal solution because it only requires a single pass through each string and uses constant-time dictionary operations to update and compare character counts.

Approach 3: Single Hash Map (Optimized Space)

Explanation:

Instead of using two dictionaries to store character frequencies, we can optimize space usage by using a single dictionary. As we iterate through both strings, we increment the character count for the first string and decrement the character count for the second string. In the end, if all values in the dictionary are zero, the strings are anagrams.

Code Implementation:

from collections import defaultdict

def isAnagram(s: str, t: str) -> bool:
    # If the lengths of the strings are not equal, they can't be anagrams
    if len(s) != len(t):
        return False

    # Create a single frequency dictionary
    char_count = defaultdict(int)

    # Increment count for characters in s and decrement for characters in t
    for i in range(len(s)):
        char_count[s[i]] += 1
        char_count[t[i]] -= 1

    # Check if all counts are zero
    return all(count == 0 for count in char_count.values())

# Test cases
print(isAnagram("anagram", "nagaram"))  # Output: True
print(isAnagram("rat", "car"))  # Output: False

Time and Space Complexity:

  • Time Complexity: O(n), since we still need to loop through both strings.
  • Space Complexity: O(n), because we use one dictionary to store the character frequencies.

Explanation:

By using a single dictionary, we can increment the count for characters in the first string and decrement the count for characters in the second string. If the final dictionary has all zero values, the strings are anagrams. This approach is optimal in terms of both time and space, as it only requires one dictionary.

Benefits:

This approach reduces the memory usage by half compared to the previous method and still maintains O(n) time complexity.

Edge Cases

When solving the Valid Anagram problem, it is important to consider the following edge cases:

  1. Different Length Strings:
    If the two strings have different lengths, they cannot be anagrams. This is an easy edge case that should be handled early in the function.

    s = "abcd"
t = "abc"
print(isAnagram(s, t))  # Output: False

  2. Empty Strings:
    Two empty strings are considered anagrams since they contain the same (zero) characters.

    s = ""
t = ""
print(isAnagram(s, t))  # Output: True

  3. Single Character Strings:
    Strings with one character can only be anagrams if they are identical.

    s = "a"
t = "a"
print(isAnagram(s, t))  # Output: True

s = "a"
t = "b"
print(isAnagram(s, t))  # Output: False

  4. Case Sensitivity:
    Depending on the problem statement, anagram checking may or may not be case-sensitive. This post assumes case-sensitive comparison.

    s = "a"
t = "A"
print(isAnagram(s, t))  # Output: False

  5. Non-Alphabetic Characters:
    If the problem allows non-alphabetic characters (such as punctuation or spaces), these should also be handled as part of the string.

    s = "a.b"
t = "b.a"
print(isAnagram(s, t))  # Output: True

Conclusion

In this post, we explored multiple ways to solve the Valid Anagram problem in Python. The sorting-based approach is straightforward but less efficient, with a time complexity of O(n log n). The hash map-based approach provides a more efficient solution, with a time complexity of O(n), and using a single hash map further optimizes the space usage. By considering edge cases, you can ensure your solution is robust and handles all possible inputs.