diff --git a/arai60/group-anagrams/step1.py b/arai60/group-anagrams/step1.py
new file mode 100644
index 0000000..4aa3d5a
--- /dev/null
+++ b/arai60/group-anagrams/step1.py
@@ -0,0 +1,20 @@
+"""
+文字列の長さは重要
+    100
+    これだと、順列を求めてとかはキビシイ
+要は、出現回数が同じなら良い
+もしくは、ソートした結果が同じでもよい
+    これがよさそう
+
+5m
+
+sorted(s)で返るのはlistじゃないことに注意
+.values()で返るのもlistじゃなくてdict_valuesという定義されたオブジェクト?
+"""
+from collections import defaultdict
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        sorted_str_to_strs = defaultdict(list)
+        for s in strs:
+            sorted_str_to_strs["".join(sorted(s))].append(s)
+        return sorted_str_to_strs.values()
diff --git a/arai60/group-anagrams/step2.py b/arai60/group-anagrams/step2.py
new file mode 100644
index 0000000..fe633d6
--- /dev/null
+++ b/arai60/group-anagrams/step2.py
@@ -0,0 +1,32 @@
+"""
+https://leetcode.com/problems/group-anagrams/solutions/3687735/beats-100-c-java-python-beginner-friendly/?envType=problem-list-v2&envId=me1nua2e
+-> anagram_map なるほど
+https://github.com/SuperHotDogCat/coding-interview/pull/13/files
+-> sorted_str_to_anagramsのほうがよさそうだな
+https://github.com/t-ooka/leetcode/pull/2/files
+-> 出現回数を持つやり方。26個のタプルを入れるのに、listからtupleに変換すればよいのはたしかに
+
+sorted_str_to_strs.values() -> list(sorted_str_to_strs.values())
+sorted_str_to_strs -> sorted_str_to_anagrams
+"""
+
+from collections import defaultdict
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        sorted_str_to_anagrams = defaultdict(list)
+        for s in strs:
+            sorted_str_to_anagrams["".join(sorted(s))].append(s)
+        return list(sorted_str_to_anagrams.values())
+
+# 出現回数を持つ
+from collections import defaultdict
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        NUM_ALPHABET = 26
+        sorted_str_to_anagrams = defaultdict(list)
+        for s in strs:
+            char_count = [0] * NUM_ALPHABET
+            for c in s:
+                char_count[ord(c) - ord('a')] += 1
+            sorted_str_to_anagrams[tuple(char_count)].append(s)
+        return list(sorted_str_to_anagrams.values())