tom4649 · tom4649 · Mar 14, 2026 · oda · Mar 14, 2026 · tom4649
diff --git a/49/memo.md b/49/memo.md
@@ -0,0 +1,11 @@
+# 49. Group Anagrams
+
+- [リンク](https://leetcode.com/problems/group-anagrams/description/)
+- 愚直にsol1を書いたが計算量はO(nllog(l))
+- histogramを用いる解法
+    - https://github.com/ichika0615/arai60/pull/11/changes#diff-fa3129c54f54ceaf0d237e449d3491c5eb44eaa4660ae11dec87df81d245cb25
+    - リストはmutableでhashableではないのでtupleに変換する必要がある
+    - 計算量はO(nl)
+    - 入力がアルファベット小文字以外のときには例外処理で対応
+- 実行時間はsol1の方が速い
+    - pythonのsortedが高速でloglが定数倍より速い？
diff --git a/49/sol1.py b/49/sol1.py
@@ -0,0 +1,10 @@
+from collections import defaultdict
+
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        anagram_sets = defaultdict(list)
+        for orig_str in strs:
+            sorted_str = sorted(orig_str)
+            anagram_sets["".join(sorted_str)].append(orig_str)
+        return list(anagram_sets.values())
diff --git a/49/sol2.py b/49/sol2.py
@@ -0,0 +1,22 @@
+from collections import defaultdict
+
+NUM_POSSIBLE_CHARS = 26
+START_CHAR = "a"
+ORD_START_CHAR = ord(START_CHAR)
+
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        def str_to_frequency(input_str):
+            frequency = [0] * NUM_POSSIBLE_CHARS
+            for input_char in input_str:
+                ord_input_char_rel = ord(input_char) - ORD_START_CHAR
+                if ord_input_char_rel >= NUM_POSSIBLE_CHARS:
+                    raise ValueError
+                frequency[ord_input_char_rel] += 1
+            return tuple(frequency)
+
+        frequency_to_strs = defaultdict(list)
+        for input_str in strs:
+            frequency_to_strs[str_to_frequency(input_str)].append(input_str)
+        return list(frequency_to_strs.values())