solution: valid anagram

Author: obzva

valid-anagram/flynn.py

@@ -0,0 +1,25 @@
+'''
+For N, length of the given strings,
+
+Time Complexity: O(N)
+- first iteration: O(N)
+- second iteration: O(N)
+
+Space Compelxity: O(N)
+- dictionaries for each strings: O(N)
+'''
+
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+
+        count_s, count_t = {}, {}
+        for i in range(len(s)):
+            count_s[s[i]] = count_s.get(s[i], 0) + 1
+            count_t[t[i]] = count_t.get(t[i], 0) + 1
+        
+        for c in count_s:
+            if c not in count_t or count_s[c] != count_t[c]:
+                return False
+        return True