0383EasyArraySortingQuicksort

Single Printer Feasibility

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A print shop owns one plotter and has a day's worth of booked jobs. jobs[i] = [start_i, end_i] means job i occupies the plotter from minute start_i up to minute end_i.

The plotter can run one job at a time. A job that ends at the same minute another one starts is fine, since the sheet is already out of the tray.

Return true if every booked job can run on the single plotter, and false if two of them would need it at the same time. The jobs are given in booking order, which need not be time order.

Examples

Example 1

Input: jobs = [[9, 12], [3, 7], [13, 16]]
Output: true

In time order the jobs occupy minutes 3-7, 9-12 and 13-16, and no two of those spans share a minute.

Example 2

Input: jobs = [[2, 8], [6, 10]]
Output: false

Minutes 6, 7 and 8 are claimed by both jobs.

Example 3

Input: jobs = [[4, 6], [6, 9], [9, 11]]
Output: true

Each job starts exactly when the previous one ends, which is allowed.

Example 4

Input: jobs = [[0, 1], [5, 9], [6, 8]]
Output: false

The second and third jobs both claim minutes 6, 7 and 8.

Constraints

0 <= jobs.length <= 10^4
jobs[i].length == 2
0 <= start_i < end_i <= 10^6

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def can_attend_meetings(jobs: list[list[int]]) -> bool:

Java

public boolean canAttendMeetings(int[][] jobs)

The intended approach

Time O(k log k)Space O(k) for the sorted copy

Put the jobs in start-time order. In that order the only way two jobs can collide is for one to start before the job just before it has ended, so a single walk over consecutive pairs settles the question. A start equal to the previous end is accepted, so the comparison is strict.

Pythonentry point: can_attend_meetings

def can_attend_meetings(jobs):
    # Time O(k log k) for the sort of k jobs.
    # Space O(k) for the sorted copy.
    ordered = sorted(jobs, key=lambda job: job[0])
    for i in range(1, len(ordered)):
        if ordered[i][0] < ordered[i - 1][1]:
            return False
    return True

The obvious version

Test every pair of jobs directly for a shared minute, using the standard overlap condition. Quadratic and needs no ordering at all.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_can_attend_meetings(jobs):
    for i in range(len(jobs)):
        for j in range(i + 1, len(jobs)):
            a_start, a_end = jobs[i]
            b_start, b_end = jobs[j]
            if a_start < b_end and b_start < a_end:
                return False
    return True

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
4 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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