Meeting Rooms
Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...,[sn,en]], determine if a person could attend all meetings.
Constraints:
- 0 <= intervals.length <= 1000
- 0 <= starti < endi <= 10^9
Examples:
Input: [[0,30],[5,10],[15,20]]
Output: false
Explanation: The person cannot attend all meetings because the meeting [5,10] overlaps with [0,30].
Input: [[5,8],[9,15]]
Output: true
Explanation: The person can attend all meetings because there are no overlapping meetings.
Solutions
Sorting and Two Pointers
First, sort the meeting intervals by their start times. Then, iterate through the sorted intervals and check if the start time of the current interval is less than the end time of the previous interval. If it is, return false because the person cannot attend all meetings. If the loop completes without finding any overlapping meetings, return true.
bool canAttendMeetings(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end(), [](const vector<int>& a, const vector<int>& b) {
return a[0] < b[0];
}
);
for (int i = 1;
i < intervals.size();
i++) {
if (intervals[i][0] < intervals[i-1][1]) return false;
}
return true;
}Follow-up:
What if the meeting intervals are not given in the format [start, end] but as separate start and end times?
