diff --git a/3.Longest-Substring-Without-Repeating-Characters/__pycache__/sol2.cpython-310.pyc b/3.Longest-Substring-Without-Repeating-Characters/__pycache__/sol2.cpython-310.pyc
new file mode 100644
index 0000000..ffd6f1d
Binary files /dev/null and b/3.Longest-Substring-Without-Repeating-Characters/__pycache__/sol2.cpython-310.pyc differ
diff --git a/3.Longest-Substring-Without-Repeating-Characters/memo.md b/3.Longest-Substring-Without-Repeating-Characters/memo.md
new file mode 100644
index 0000000..12c692c
--- /dev/null
+++ b/3.Longest-Substring-Without-Repeating-Characters/memo.md
@@ -0,0 +1,24 @@
+# 3. Longest Substring Without Repeating Characters
+
+- sol1.py: 自力で解いたが、テストの途中を print するなどデバッグして答えを合わせた
+
+コメント集
+- https://github.com/olsen-blue/Arai60/pull/49#discussion_r2005295464
+> seen_char_to_index.get(s[right], -1) と使えば、条件分岐を回避できますね。
+
+
+
+知らなかった（勉強したことはある気がする）が、初期の実装がこのアルゴリズムになっていた
+- 尺取法
+    - https://qiita.com/drken/items/ecd1a472d3a0e7db8dce
+- sliding window
+    - https://qiita.com/rumblekat03/items/4edd9dd4607280c994d1
+
+
+https://github.com/mamo3gr/arai60/blob/3_longest-substring-without-repeating-characters/3_longest-substring-without-repeating-characters/memo.md
+
+> char_to_index.get(c, -1) として条件分岐をなくす (step2). コードとしてはすっきりするけど、読み手からしたらシミュレートの手間が増えそう。
+
+という考えもあるのか。
+
+sol2.py: 他の人のコードとコメントを参考にもう一度書き直す
diff --git a/3.Longest-Substring-Without-Repeating-Characters/sol1.py b/3.Longest-Substring-Without-Repeating-Characters/sol1.py
new file mode 100644
index 0000000..ebff081
--- /dev/null
+++ b/3.Longest-Substring-Without-Repeating-Characters/sol1.py
@@ -0,0 +1,21 @@
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        if not s:
+            return 0
+        seen_char_to_idx = {}
+        start = 0
+        max_length = 0
+        last = start + 1
+        seen_char_to_idx[s[start]] = start
+        while start < len(s):
+            while last < len(s) and s[last] not in seen_char_to_idx:
+                seen_char_to_idx[s[last]] = last
+                last += 1
+            max_length = max(max_length, last - start)
+            if last >= len(s):
+                break
+            start = max(seen_char_to_idx[s[last]] + 1, start)
+            seen_char_to_idx[s[last]] = last
+            last += 1
+
+        return max_length
diff --git a/3.Longest-Substring-Without-Repeating-Characters/sol2.py b/3.Longest-Substring-Without-Repeating-Characters/sol2.py
new file mode 100644
index 0000000..b5b0fc9
--- /dev/null
+++ b/3.Longest-Substring-Without-Repeating-Characters/sol2.py
@@ -0,0 +1,12 @@
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        last_index = {}
+        start = 0
+        max_length = 0
+        for end, ch in enumerate(s):
+            prev = last_index.get(ch, -1)
+            if prev >= start:
+                start = prev + 1
+            last_index[ch] = end
+            max_length = max(max_length, end - start + 1)
+        return max_length