Is Meeting Rooms easy or hard?

Meeting Rooms is classified as an Easy problem on LeetCode with an acceptance rate around 59%. The challenge mainly involves recognizing that sorting intervals simplifies overlap detection, a pattern that appears in many interval-based interview questions.

Meeting Rooms Python/Java solution

In Python or Java, sort the intervals using the built-in sort function with the start time as the key. Then iterate from the second interval onward and check if the current start is less than the previous end. If it is, return false; otherwise continue until the array ends.

How to solve Meeting Rooms in O(n)?

A true O(n) solution is not possible for arbitrary unsorted intervals because you must first order them by time to reliably detect overlaps. Sorting requires O(n log n). Once sorted, the overlap detection step itself runs in O(n) with a simple linear scan.

What is the best approach for Meeting Rooms?

The best approach sorts the meeting intervals by start time and then performs a single linear scan to detect overlaps. After sorting, compare each meeting's start time with the previous meeting's end time. If any start time is smaller than the previous end, the meetings overlap. This runs in O(n log n) time and O(1) extra space.

Is Meeting Rooms asked at Google/Amazon/Meta?

Meeting Rooms is a common interval scheduling question used in interviews at companies like Google, Amazon, Meta, and other tech firms. It tests understanding of interval overlap detection, sorting strategies, and reasoning about time ranges.

What data structure is used in Meeting Rooms?

The problem primarily uses arrays to store meeting intervals. The key technique is sorting the array based on start times, followed by a sequential scan to compare adjacent intervals and detect overlaps.

What is the time complexity of Meeting Rooms?

The optimal solution runs in O(n log n) time due to sorting the intervals by their start times. After sorting, checking for overlaps requires only a single O(n) pass through the array. The brute force alternative takes O(n²) time because it compares every pair of intervals.

Description Solution Video 8:16

252. Meeting Rooms

Easy59.4% acceptance10 companiesPremium elsewhere, free here

Given an array of meeting time intervals where intervals[i] = [start_i, end_i], determine if a person could attend all meetings.

Example 1:

Input: intervals = [[0,30],[5,10],[15,20]]
Output: false

Example 2:

Input: intervals = [[7,10],[2,4]]
Output: true

Constraints:

0 <= intervals.length <= 10⁴
intervals[i].length == 2
0 <= start_i < end_i <= 10⁶

Asked at 10 companies

Apple Tiktok Oracle Palo Alto Networks Microsoft Meta

Google Amazon Bloomberg Uber

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Array Sorting

8:16

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252. Meeting Rooms

Asked at 10 companies

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