# Meeting Rooms

**Easy**

Given an array of meeting time intervals consisting of start and end times`[[s1,e1],[s2,e2],...]`(si< ei), determine if a person could attend all meetings.

**Example 1:**

```
Input:
[[0,30],[5,10],[15,20]]
Output:
 false
```

**Example 2:**

```
Input:
 [[7,10],[2,4]]

Output:
 true
```

## Solution & Analysis

> The idea here is to sort the meetings by **starting time**. Then, go through the meetings one by one and make sure that each meeting **ends before the next one starts**.

* **Time complexity** : O(nlogn). The time complexity is dominated by sorting. Once the array has been sorted, only O(n) time is taken to go through the array and determine if there is any overlap.
* **Space complexity** : O(1). Since no additional space is allocated.

**先排序，再遍历**

Sort, then compare last.end vs. current.start; similar to Merge Intervals

```java
/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class Solution {
    public boolean canAttendMeetings(Interval[] intervals) {
        Arrays.sort(intervals, new Comparator<Interval>() {
           public int compare(Interval i1, Interval i2) {
               return i1.start - i2.start;
           } 
        });
        Interval last = null;
        for (Interval i: intervals) {
            if (last != null && i.start < last.end) {
                return false;
            }
            last = i;
        }
        return true;
    }
}
```

Sorting

```java
public boolean canAttendMeetings(Interval[] intervals) {
    Arrays.sort(intervals, new Comparator<Interval>() {
        public int compare(Interval i1, Interval i2) {
            return i1.start - i2.start;
        }        
    });
    for (int i = 0; i < intervals.length - 1; i++) {
        if (intervals[i].end > intervals[i + 1].start) return false;
    }
    return true;
}
```

Other implementation

Sorting - throw expection

```java
private boolean canAttendMeetings(Interval[] intervals) {
    try {
        Arrays.sort(intervals, new IntervalComparator());
    } catch (Exception e) {
        return false;
    }
    return true;
}

private class IntervalComparator implements Comparator<Interval> {
    @Override
    public int compare(Interval o1, Interval o2) {
        if (o1.start < o2.start && o1.end <= o2.start)
            return -1;
        else if (o1.start > o2.start && o1.start >= o2.end)
            return 1;
        throw new RuntimeException();
    }
}
```

Java 8

```java
public boolean canAttendMeetings(Interval[] intervals) {
    // Sort the intervals by start time
    Arrays.sort(intervals, (x, y) -> x.start - y.start);
    for (int i = 1; i < intervals.length; i++)
        if (intervals[i-1].end > intervals[i].start)
            return false;
    return true;
}
```

## Reference

<https://leetcode.com/problems/meeting-rooms/discuss/67786/AC-clean-Java-solution>


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