LeetCode 242 Valid Anagram in Python | Easy Coding Tutorial for Beginners
Solve LeetCode 242 "Valid Anagram" in Python with this beginner-friendly coding tutorial! This problem asks you to determine if two strings are anagrams of each other—same characters, same frequencies, different order (e.g., s = "anagram", t = "nagaram", return true). We’ll start with a sorting solution, then explore a more efficient approach using a frequency dictionary. Perfect for Python learners, coding beginners, or anyone prepping for coding interviews!
🔍 *What You'll Learn:*
Understanding LeetCode 242’s anagram requirements
Using sorting to compare strings
Optimizing with a frequency dictionary for better performance
Testing with example cases
💻 *Code Used in This Video:*
class Solution(object):
def isAnagram(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
if len(s) != len(t):
return False
return sorted(s) == sorted(t)
Test cases
solution = Solution()
Test case 1: Valid anagram
print(solution.isAnagram("anagram", "nagaram")) # Output: True
Same characters, same frequencies, different order
Test case 2: Not an anagram
print(solution.isAnagram("rat", "car")) # Output: False
Different characters
Test case 3: Different lengths
print(solution.isAnagram("hello", "hell")) # Output: False
Different lengths, cannot be anagrams
Test case 4: Empty strings
print(solution.isAnagram("", "")) # Output: True
Both empty, so they are anagrams
Optimized: Using a frequency dictionary
from collections import Counter
class SolutionOptimized(object):
def isAnagram(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
if len(s) != len(t):
return False
return Counter(s) == Counter(t)
solution_opt = SolutionOptimized()
print("\nOptimized solution:")
print(solution_opt.isAnagram("anagram", "nagaram")) # Output: True
print(solution_opt.isAnagram("rat", "car")) # Output: False
🌟 *Why Solve LeetCode 242?*
This problem is a great introduction to string manipulation and frequency counting in Python, a common topic in coding interviews! The sorting solution has a time complexity of O(n log n), but the optimized solution using a frequency dictionary reduces it to O(n) with O(1) space complexity (since the character set is fixed). We’ll explore both methods to check for anagrams efficiently. Master this, and you’ll be ready for more advanced LeetCode challenges!
📚 *Who’s This For?*
Python beginners learning coding
Coding enthusiasts tackling LeetCode problems
Developers prepping for technical interviews
👍 Like, subscribe, and comment: What LeetCode problem should we solve next? Next up: More LeetCode string problems—stay tuned!
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