1096MediumArraySorting

Days the Workshop Stays Free

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Trains the technique from

LeetCode 3169Count Days Without Meetings

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 workshop takes bookings over days days numbered 1 through days. bookings[i] = [first, last] reserves the workshop for every day from first to last, both ends included. Bookings may overlap, sit inside one another or arrive in any order.

Return how many of the days days carry no booking at all.

Examples

Example 1

Input: days = 12, bookings = [[5, 7], [1, 3], [9, 10]]
Output: 4

Days 1 to 3, 5 to 7 and 9 to 10 are reserved. Day 4, day 8, day 11 and day 12 are left.

Example 2

Input: days = 20, bookings = [[5, 10], [6, 8], [7, 7]]
Output: 14

The second and third bookings sit wholly inside the first, so only days 5 to 10 are reserved and the other fourteen days stay free.

Example 3

Input: days = 1, bookings = [[1, 1]]
Output: 0

The workshop's single day is booked.

Constraints

1 <= days <= 10^9
1 <= bookings.length <= 10^5
bookings[i].length == 2
1 <= bookings[i][0] <= days
1 <= bookings[i][1] <= days
bookings[i][0] <= bookings[i][1]

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 count_days(days: int, bookings: list[list[int]]) -> int:

Java

public int countDays(int days, int[][] bookings)

The intended approach

Time O(m log m) for m bookings.Space O(m) for the sorted order.

The day count runs to a billion, so the days cannot be visited one at a time. Sort the bookings by the day they start and sweep through them holding one number: the last day known to be reserved so far. When a booking starts more than one day past that mark, the days strictly between are free and can be added in one stroke. Then push the mark forward to this booking's last day, but only if that is later: a booking wholly inside an earlier one must not drag the mark back, which is what makes nesting and overlap both fall out for free. Once every booking is handled, whatever lies between the mark and the final day is free as well. Sorting dominates the cost, and the sweep itself touches each booking once.

Pythonentry point: count_days

def count_days(days, bookings):
    order = sorted(bookings, key=lambda span: span[0])
    free = 0
    reach = 0
    for span in order:
        if span[0] > reach + 1:
            free += span[0] - reach - 1
        if span[1] > reach:
            reach = span[1]
    if days > reach:
        free += days - reach
    return free

The obvious version

Keep a flag for every single day, walk each booking and mark every day it covers, then count the days left unmarked. Costs the day count plus the total length of the bookings, which is hopeless once the day count runs into the millions.

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_count_days(days, bookings):
    busy = [False] * (days + 1)
    for span in bookings:
        for day in range(span[0], span[1] + 1):
            busy[day] = True
    free = 0
    for day in range(1, days + 1):
        if not busy[day]:
            free += 1
    return free

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 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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