DSA TrainerMeeting Rooms
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 TrueO(n²) — all pairs.
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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