From 39f1241ded793beaccea394ed28321bb04aeb36d Mon Sep 17 00:00:00 2001
From: Sam2022moon <104721736+Sam2022moon@users.noreply.github.com>
Date: Mon, 20 May 2024 15:49:38 +0900
Subject: [PATCH] solved group anagram

---
 group-anagrams/samthekorean.py | 15 +++++++++++++++
 1 file changed, 15 insertions(+)
 create mode 100644 group-anagrams/samthekorean.py

diff --git a/group-anagrams/samthekorean.py b/group-anagrams/samthekorean.py
new file mode 100644
index 000000000..4ea2fcf08
--- /dev/null
+++ b/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()