valid-anagram

Author: changhyumm

valid-anagram/changhyumm.py

@@ -0,0 +1,26 @@
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+        # sorted_s = sorted(s)
+        # sorted_t = sorted(t)
+        # for i in range(len(s)):
+        #     if sorted_s[i] != sorted_t[i]:
+        #         return False
+        # return True
+        # 시간복잡도가 O(nlog(n))
+
+        # Hashmap 사용하는 방법으로 변경
+        counter = {}
+        for char in s:
+            counter[char] = counter.get(char, 0) + 1
+        for char in t:
+            if char in counter and counter[char] != 0:
+                counter[char] -= 1
+            elif char in counter and counter[char] == 0:
+                return False
+            else:
+                return False
+        # 시간복잡도 O(n)
+        # 공간복잡도 O(n)
+        return True