Merge pull request #727 from yolophg/main

[Helena] Week 2

Author: SamTheKorean

valid-anagram/yolophg.py

@@ -0,0 +1,31 @@
+# Time Complexity: O(n)  
+# Go through both strings once to count and compare characters.
+
+# Space Complexity: O(1)  
+# The dictionary stores at most 26 letters, so space usage is constant.
+
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        char_count = {}
+        
+        # count frequencies of characters in string s
+        for char in s:
+            if char in char_count:
+                char_count[char] += 1
+            else:
+                char_count[char] = 1
+
+        # go through 't' and decrease the count
+        for char in t:
+            # if the character isn't in the dictionary, it's not an anagram
+            if char in char_count:
+                char_count[char] -= 1
+            else:
+                return False
+
+        # check if all the counts are zero
+        for val in char_count.values():
+            if val != 0:
+                return False
+        
+        return True