0093HardArrayBinary SearchDynamic ProgrammingSorting

Studio Booking Payout

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Trains the technique from

LeetCode 1235Maximum Profit in Job Scheduling

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 recording studio has a stack of booking requests to answer. Request i wants the room from hour start[i] until hour finish[i] and pays payout[i] if it is accepted.

The room holds one session at a time, so two accepted requests may not overlap. Requests that merely touch are fine: a request that ends at hour t can be followed by one that begins at hour t, because the engineer resets the room instantly.

Accept any collection of requests you like, subject to that rule, and return the largest total payout obtainable. Accepting nothing is allowed but never necessary, since every payout is positive.

Examples

Example 1

Input: start = [1, 3], finish = [3, 5], payout = [6, 6]
Output: 12

The second request begins on the hour the first one ends, so both fit and the studio collects everything.

Example 2

Input: start = [1, 1, 3], finish = [10, 3, 6], payout = [20, 12, 12]
Output: 24

Two shorter back-to-back sessions beat the single long one that pays 20 and blocks the whole day.

Example 3

Input: start = [1, 2, 3], finish = [8, 7, 9], payout = [5, 9, 3]
Output: 9

Every pair of requests overlaps here, so only the best paying one can be taken.

Constraints

1 <= start.length == finish.length == payout.length <= 5 * 10^4
1 <= start[i] < finish[i] <= 10^9
1 <= payout[i] <= 10^4

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 max_payout(start: list[int], finish: list[int], payout: list[int]) -> int:

Java

public int maxPayout(int[] start, int[] finish, int[] payout)

The intended approach

Time O(n log n)Space O(n)

Sort the requests by the hour they release the room, then sweep them while keeping a staircase of (release hour, best payout achievable by then) whose payouts only ever increase. Accepting a request means binary searching that staircase for the latest entry whose release hour is at or before the request's start hour and adding the fee to it; the staircase only grows when the result beats the current best, which keeps it sorted for the next search.

Pythonentry point: max_payout

def max_payout(start, finish, payout):
    from bisect import bisect_right
    bookings = sorted(zip(finish, start, payout))
    edges = [0]
    best = [0]
    for end_hour, start_hour, fee in bookings:
        idx = bisect_right(edges, start_hour) - 1
        candidate = best[idx] + fee
        if candidate > best[-1]:
            edges.append(end_hour)
            best.append(candidate)
    return best[-1]

The obvious version

Enumerate every collection of requests, reject the ones with a real overlap, and keep the richest survivor.

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_max_payout(start, finish, payout):
    n = len(start)
    best = 0
    for mask in range(2 ** n):
        picked = [i for i in range(n) if (mask >> i) % 2 == 1]
        picked.sort(key=lambda i: start[i])
        clash = False
        for k in range(len(picked) - 1):
            if finish[picked[k]] > start[picked[k + 1]]:
                clash = True
                break
        if clash:
            continue
        total = 0
        for i in picked:
            total += payout[i]
        if total > best:
            best = total
    return best

How this problem was checked

Reference solution executed against 9 test cases — 9 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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