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What it teaches: Sort by start and check neighbours: any overlap means one person can't attend everything.
Practise it on judges as “Meeting Rooms”.
Given meeting times [start, end], can one person attend all of them? A meeting ending at time t doesn't clash with one starting at t.
Example 1
Input: [[0,30],[5,10],[15,20]] Output: false
Example 2
Input: [[7,10],[2,4]] Output: true
Constraints
Pattern clues in the wording
These clues point to Merge Intervals: Sort intervals by start; each one either overlaps the last merged interval (extend it) or starts a new one.
Stuck? Take one hint at a time
import java.util.*;
class Solution {
public boolean canAttendMeetings(int[][] intervals) {
return true;
}
}Write your solution locally or in your editor for now. The in-browser runner (Java first, then Python, C++ and more) will run these tests right here.
Test cases
| # | Input | Expected |
|---|---|---|
| 1 | intervals = [[0,30],[5,10],[15,20]] | false |
| 2 | intervals = [[7,10],[2,4]] | true |
+ 1 hidden test the code runner will check
From slow to fast
Sort by start; if any meeting starts before the previous one ends, return false.
import java.util.Arrays;
class Solution {
public boolean canAttendMeetings(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
for (int i = 1; i < intervals.length; i++)
if (intervals[i][0] < intervals[i - 1][1]) return false;
return true;
}
}Verdict: Straightforward.
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How many rooms would be needed instead?