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: [[0,20],[5,10],[20,30]]
Output: false
Example 2:
Input: [[4,6],[7,10]
Output: true
Solution:
We will sort by end time of all the meetings and then check if there is a case where one meeting starts before the previous meeting stops. If such a case exists, then it is not possible to attend all the meetings. Otherwise, it is possible to attend all the meetings
Implementation:
bool canAttendMeetings(vector<Interval>& intervals) {
sort(intervals.begin(), intervals.end(), [](const Interval& i1, const Interval& i2) -> bool {
return i1.start < i2.start;
});
for (size_t i = 1; i < intervals.size(); i++) {
if (intervals[i - 1].end > intervals[i].start) {
return false;
}
}
return true;
}
class Solution {
public boolean canAttendMeetings(int[][] intervals) {
Arrays.sort(intervals, new Comparator<int[]>() {
public int compare(int[] i1, int[] i2) {
return i1[0] - i2[0];
}
});
for (int i = 0; i < intervals.length - 1; i++) {
if (intervals[i][1] > intervals[i + 1][0])
return false;
}
return true;
}
}
def canAttendMeetings(intervals):
intervals.sort(key=lambda a: a.start)
for i in range(len(intervals)-1):
if intervals[i].end > intervals[i+1].start:
return False
return True
Complexity Analysis:
- Time Complexity: O(nlogn).
- Space Complexity: O(1).
