What is the time complexity of the brute-force approach to Meeting Rooms?

The brute-force approach runs in O(n²) time and O(1) space.

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Meeting Rooms

Q: What pattern does Meeting Rooms use?

Meeting Rooms is a Greedy problem. The key question to ask is: After sorting by start time, why is it enough to compare only adjacent intervals? Could a non-adjacent pair overlap while adjacent ones don't?

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Given an array of meeting time intervals [start, end], determine if a person can attend all meetings (no two intervals overlap).

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Problem

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

Input

An array of intervals `intervals` where `intervals[i] = [start, end]`.

Output

`true` if a person can attend all meetings, `false` if any two meetings overlap.

Examples

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

Output: false

[0,30] overlaps with [5,10] and [15,20].

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

Output: true

[2,4] ends before [7,10] starts.

The brute-force approach

Check every pair of intervals. If any two overlap, return false.

for i in range(n):
    for j in range(i+1, n):
        if intervals[i][0] < intervals[j][1] and intervals[j][0] < intervals[i][1]:
            return False
return True

O(n²) — all pairs.

Time: O(n²)Space: O(1)

Spotting the pattern

This is a Greedy problem. The key question to ask yourself:

After sorting by start time, why is it enough to compare only adjacent intervals? Could a non-adjacent pair overlap while adjacent ones don't?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.

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