4. Meeting room

Given an array of meeting time interval objects consisting of start and end times [[start_1,end_1],[start_2,end_2],...] (start_i < end_i), determine if a person could add all meetings to their schedule without any conflicts.

Example 1:

Input: intervals = [(0,30),(5,10),(15,20)]
 
Output: false

Explanation:

• (0,30) and (5,10) will conflict
• (0,30) and (15,20) will conflict

Example 2:

Input: intervals = [(5,8),(9,15)]
 
Output: true

Note:

• (0,8),(8,10) is not considered a conflict at 8

Constraints:

• 0 <= intervals.length <= 500
• 0 <= intervals[i].start < intervals[i].end <= 1,000,000

Approach - sort

• Just sort by start and then check previous end current start
• Time & Space Complexity
  Time complexity: O(nlog⁡n)
  Space complexity: O(1) or O(n) depending on the sorting algorithm.

/**
 * Definition of Interval:
 * public class Interval {
 *     public int start, end;
 *     public Interval(int start, int end) {
 *         this.start = start;
 *         this.end = end;
 *     }
 * }
 */
 
public class Solution {
    public boolean canAttendMeetings(List<Interval> intervals) {
        Collections.sort(intervals, Comparator.comparingInt(i -> i.start));
 
        for (int i = 1; i < intervals.size(); i++) {
            Interval i1 = intervals.get(i - 1);
            Interval i2 = intervals.get(i);
 
            if (i1.end > i2.start) {
                return false;
            }
        }
 
        return true;
    }
}