Meeting Rooms

JavaView on GFG

Intuition

Check if a person can attend all meetings (no overlaps). Sort by start, check for overlaps.

Algorithm

1Sort intervals by start. For i in 1..n-1: if intervals[i].start < intervals[i-1].end: overlap, return false.

Common Pitfalls

•Same as LC 252. Sort by start time. Overlap if next start < previous end.

Meeting Rooms.java

Java

// Approach: Sort intervals by start time. Check if any interval starts before the previous one ends.
// Time: O(n log n) Space: O(1)
import java.util.*;

class Solution {
    static boolean canAttend(int[][] arr) {
        Arrays.sort(arr, (a, b) -> a[0] - b[0]);
        int end = arr[0][1];
        for (int i = 1; i < arr.length; i++) {
            if (end > arr[i][0]) {
                return false;
            }
            end = arr[i][1];
        }
        return true;
    }
}

Was this solution helpful?