add solution: valid-anagram

Author: dusunax

valid-anagram/dusunax.py

@@ -0,0 +1,44 @@
+'''
+# Leetcode 242. Valid Anagram
+
+use `Counter` to (1)compare the frequency of characters in both strings, and (2)try to compare the frequency more efficiently. 🔍
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### A. use frequency object
+
+#### TC is O(n):
+- iterating through the strings just once to compare the frequency of characters. O(n)
+
+#### SC is O(n):
+- creating a new string `converted_s` to store the 
+
+### B. use Counter more efficiently
+
+#### TC is O(n):
+- iterating through the strings just once to compare the frequency of characters. O(n)
+
+#### SC is O(n):
+- creating a new string `converted_s` to store the 
+'''
+class Solution:
+    def isAnagramA(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+
+        frequency = Counter(s) # SC: O(n)
+
+        for char in t: # TC: O(n)
+            if char not in frequency or frequency[char] == 0: # TC: O(1)
+                return False
+            frequency[char] -= 1
+
+        return True
+    
+    def isAnagramB(self, s: str, t: str) -> bool:
+        return Counter(s) == Counter(t) # TC: O(n), SC: O(n)