diff --git a/252_meeting_rooms_easy/README.md b/252_meeting_rooms_easy/README.md
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+# 問題へのリンク
+[Meeting Rooms - LeetCode](https://leetcode.com/problems/meeting-rooms/)
+
+# 言語
+Python
+
+# 問題の概要
+
+ある人が複数の会議に出席できるかを判定する問題である。
+
+* 各会議は時間帯 `[starti, endi]` で表される。
+* 入力は、こうした会議の時間帯のリスト `intervals` で与えられる。
+* **すべての会議に出席するには、会議同士が重ならない必要がある。**
+
+#### 例1:
+
+```
+入力: [[0,30],[5,10],[15,20]]
+出力: false（0〜30と5〜10が重なるため）
+```
+
+#### 例2:
+
+```
+入力: [[7,10],[2,4]]
+出力: true（会議が重ならないため）
+```
+
+#### 制約:
+
+* 会議数は 0〜10,000 件
+* 各会議の時間帯は `[starti, endi]`（開始 < 終了、かつ0以上1,000,000以下）
+
+
+# 自分の解法
+`intervals` の時間帯を開始時間でソートし、隣接する会議の終了時間と開始時間を比較する方法。ひとつでも重なっている会議があれば、`False` を返す。
+- 時間計算量：`O(nlogn)`
+- 空間計算量：`O(1)`
+
+## step2
+
+- `start`, `end` の変数名を `start_time`, `end_time` に変更して、より明確にした。
+
+# 別解・模範解答
+`intervals` からすべての時間帯のペアを取り出し、そのペアが重なるかどうかを確認する方法。
+
+- 時間計算量：`O(n^2)`（すべてのペアを確認するため）
+- 空間計算量：`O(1)`
+
+# 次に解く問題の予告
+- [Largest Component Size by Common Factor - LeetCode](https://leetcode.com/problems/largest-component-size-by-common-factor/description/)
+- [Longest Consecutive Sequence - LeetCode](https://leetcode.com/problems/longest-consecutive-sequence/description/)
diff --git a/252_meeting_rooms_easy/another_solution.py b/252_meeting_rooms_easy/another_solution.py
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+#
+# @lc app=leetcode id=252 lang=python3
+#
+# [252] Meeting Rooms
+#
+
+
+# @lc code=start
+TIME_POINT_START = 1
+TIME_POINT_END = -1
+
+
+class Solution:
+    def canAttendMeetings(self, intervals: list[list[int]]) -> bool:
+        time_points: list[tuple[int, int]] = []
+        for start_time, end_time in intervals:
+            time_points.append((start_time, TIME_POINT_START))
+            time_points.append((end_time, TIME_POINT_END))
+
+        time_points.sort()
+        # time_points must have start time and end time in alternating order after sorting
+        # if not, then there are overlapping intervals
+        previous_time_type = TIME_POINT_END
+        for _, time_type in time_points:
+            if time_type == previous_time_type:
+                return False
+            previous_time_type = time_type
+        return True
+
+
+# @lc code=end
diff --git a/252_meeting_rooms_easy/bruteforce.py b/252_meeting_rooms_easy/bruteforce.py
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+#
+# @lc app=leetcode id=252 lang=python3
+#
+# [252] Meeting Rooms
+#
+
+
+# @lc code=start
+class Solution:
+    def canAttendMeetings(self, intervals: list[list[int]]) -> bool:
+
+        def are_overlapping(
+            start_time1: int,
+            end_time1: int,
+            start_time2: int,
+            end_time2: int,
+        ) -> bool:
+            if start_time1 < end_time1 <= start_time2 < end_time2:
+                return False
+            elif start_time2 < end_time2 <= start_time1 < end_time1:
+                return False
+            return True
+
+        for i in range(len(intervals) - 1):
+            for j in range(i + 1, len(intervals)):
+                if are_overlapping(*intervals[i], *intervals[j]):
+                    return False
+        return True
+
+
+# @lc code=end
diff --git a/252_meeting_rooms_easy/step1.py b/252_meeting_rooms_easy/step1.py
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+#
+# @lc app=leetcode id=252 lang=python3
+#
+# [252] Meeting Rooms
+#
+
+
+# @lc code=start
+
+
+class Solution:
+    def canAttendMeetings(self, intervals: list[list[int]]) -> bool:
+        intervals.sort()
+        previous_end = 0
+        for start, end in intervals:
+            if start < previous_end:
+                return False
+            previous_end = end
+        return True
+
+
+# @lc code=end
diff --git a/252_meeting_rooms_easy/step2.py b/252_meeting_rooms_easy/step2.py
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+++ b/252_meeting_rooms_easy/step2.py
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+#
+# @lc app=leetcode id=252 lang=python3
+#
+# [252] Meeting Rooms
+#
+
+
+# @lc code=start
+class Solution:
+    def canAttendMeetings(self, intervals: list[list[int]]) -> bool:
+        intervals.sort()
+        previous_end_time = 0
+        for start_time, end_time in intervals:
+            if start_time < previous_end_time:
+                return False
+            previous_end_time = end_time
+        return True
+
+
+# @lc code=end
diff --git a/252_meeting_rooms_easy/step3.py b/252_meeting_rooms_easy/step3.py
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+#
+# @lc app=leetcode id=252 lang=python3
+#
+# [252] Meeting Rooms
+#
+
+
+# @lc code=start
+
+
+class Solution:
+    def canAttendMeetings(self, intervals: list[list[int]]) -> bool:
+        intervals.sort()
+        previous_end_time = 0
+        for start_time, end_time in intervals:
+            if start_time < previous_end_time:
+                return False
+            previous_end_time = end_time
+        return True
+
+
+# @lc code=end