Check if a person can attend all meetings?

Companies:

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).
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