LeetCode 252: Meeting Rooms Solution

Problem Explanation

Explanation:

The problem "Meeting Rooms" asks us to determine if a person can attend all meetings without overlapping. We are given a list of intervals where each interval represents a meeting.

To solve this problem, we can follow these steps:

Sort the intervals based on their start times.
Iterate through the sorted intervals and check if the current interval's start time is less than the previous interval's end time. If true, it means there is an overlap, and the person cannot attend all meetings.

The time complexity of this approach is O(nlogn) due to sorting, where n is the number of intervals. The space complexity is O(1) as we are not using any extra space.

Solution Code

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;
    }
}

Frequently Asked Questions

How to solve LeetCode 252 (Meeting Rooms)?

This page provides optimized solutions for LeetCode problem 252 (Meeting Rooms) in Java, C++, and Python, along with a detailed explanation and an interactive code editor to test your code.

What is the time complexity of LeetCode 252 (Meeting Rooms)?

The time complexity for LeetCode 252 (Meeting Rooms) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

Can I run code for LeetCode 252 on DevExCode?

Yes, DevExCode provides an interactive code editor where you can write, test, and run your code for LeetCode 252 in Java, C++, or Python.

252. Meeting Rooms