solved group anagram

group-anagrams/samthekorean.py

@@ -0,0 +1,15 @@
+# TC : O(nwlog(w))
+# reason : n being the number of words and w being the length of each word, therefore the total time complexity is n times wlog(w) (time complexity for sorting each word)
+# SC : O(w*n)
+from collections import defaultdict
+
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        anagrams = defaultdict(list)
+
+        # Use sorted word as a string and append it with the original word so the word containing same charactors with the same number of existence can be in the same group
+        for word in strs:
+            anagrams[str(sorted(word))].append(word)
+
+        return anagrams.values()