From 18738146e5cc9d9c15d9471775847dcf18652bba Mon Sep 17 00:00:00 2001
From: jonghoonpark <dev@jonghoonpark.com>
Date: Tue, 23 Jul 2024 04:50:17 +0900
Subject: [PATCH] =?UTF-8?q?add=20post=20'(Leetcode)=20253=20-=20=20Meeting?=
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 _posts/2024-05-14-leetcode-252.md |  2 +-
 _posts/2024-07-22-leetcode-253.md | 83 +++++++++++++++++++++++++++++++
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diff --git a/_posts/2024-05-14-leetcode-252.md b/_posts/2024-05-14-leetcode-252.md
index 10b5576..e3c2701 100644
--- a/_posts/2024-05-14-leetcode-252.md
+++ b/_posts/2024-05-14-leetcode-252.md
@@ -1,6 +1,6 @@
 ---
 layout: post
-title: (Leetcode) 252 - Meeting Schedule
+title: (Leetcode) 252 - Meeting Room 풀이 (Meeting Schedule)
 categories: [스터디-알고리즘]
 tags:
   [자바, java, 리트코드, Leetcode, 알고리즘, algorithm, list, 리스트, neetcode]
diff --git a/_posts/2024-07-22-leetcode-253.md b/_posts/2024-07-22-leetcode-253.md
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--- /dev/null
+++ b/_posts/2024-07-22-leetcode-253.md
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+---
+layout: post
+title: (Leetcode) 253 -  Meeting Room ii 풀이 (Meeting Schedule ii)
+categories: [스터디-알고리즘]
+tags: [자바, java, 리트코드, Leetcode, 알고리즘, interval, array, sorting]
+date: 2024-07-22 19:30:00 +0900
+toc: true
+---
+
+기회가 되어 [달레님의 스터디](https://github.com/DaleStudy/leetcode-study)에 참여하여 시간이 될 때마다 한문제씩 풀어보고 있다.
+
+[https://neetcode.io/practice](https://neetcode.io/practice)
+
+---
+
+- 문제
+  - 유료: [https://leetcode.com/problems/meeting-rooms-ii/](https://leetcode.com/problems/meeting-rooms-ii/)
+  - 무료: [https://neetcode.io/problems/meeting-schedule-ii](https://neetcode.io/problems/meeting-schedule-ii)
+
+## 내가 작성한 풀이
+
+지난번에 풀었던 [meeting room](https://algorithm.jonghoonpark.com/2024/05/14/leetcode-252) 문제의 심화버전이다.
+해당 문제에서 사용했던 풀이를 재활용하였다.
+
+먼저 intervals를 정렬한다.
+이후 반복문을 통해 interval을 추가해나가는데, 만약 겹치는 시간대가 있을 경우 새로운 날짜를 생성한다.
+
+```java
+public class Solution {
+    public int minMeetingRooms(List<Interval> intervals) {
+        intervals = intervals.stream().sorted(Comparator.comparingInt(o -> o.start)).toList();
+
+        List<List<Interval>> days = new ArrayList<>();
+
+        for(Interval interval : intervals) {
+            boolean added = false;
+            for (List<Interval> day : days) {
+                day.add(interval);
+                if (canAttendMeetings(day)) {
+                    added = true;
+                    break;
+                }
+                day.remove(day.size() - 1);
+            }
+
+            if (!added) {
+                List<Interval> newDay = new ArrayList<>();
+                newDay.add(interval);
+                days.add(newDay);
+            }
+        }
+
+        return days.size();
+    }
+
+    public boolean canAttendMeetings(List<Interval> intervals) {
+        for (int i = 0; i < intervals.size() - 1; i++) {
+            if(intervals.get(i).end > intervals.get(i+1).start) {
+                return false;
+            }
+        }
+        return true;
+    }
+}
+
+class Interval {
+    public int start, end;
+
+    public Interval(int start, int end) {
+        this.start = start;
+        this.end = end;
+    }
+
+    @Override
+    public String toString() {
+        return "{" + start + ", " + end + "}";
+    }
+}
+```
+
+### TC, SC
+
+days의 길이를 m 이라고 했을 때, 시간 복잡도는 `O(n^2 * m)` 공간 복잡도는 `O(n)` 이다. m 은 최악의 경우 n 이 될 수 있다.