1172HardArrayBinary SearchDynamic ProgrammingSorting

Taking at Most k Bookings

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Trains the technique from

LeetCode 1751Maximum Number of Events That Can Be Attended II

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 hall has bookings on offer. bookings[i] = [first, last, fee] runs from day first to day last, both days included, and pays fee.

Take at most k of them, and no two taken bookings may share a day.

Return the largest total fee that can be earned.

Examples

Example 1

Input: bookings = [[1, 2, 4], [2, 3, 10]], k = 2
Output: 10

The two bookings share day 2, so only one may be taken and the larger fee wins.

Example 2

Input: bookings = [[1, 10, 5], [2, 3, 4], [5, 6, 4]], k = 2
Output: 8

The long booking pays 5 on its own, while the two short ones share no day and pay 4 each.

Example 3

Input: bookings = [[1, 2, 1], [3, 4, 1], [5, 6, 1], [7, 8, 1]], k = 2
Output: 2

The four bookings share no days at all, but only two of them may be taken.

Constraints

1 <= k <= bookings.length
1 <= k * bookings.length <= 10^6
bookings[i].length == 3
1 <= bookings[i][0] <= bookings[i][1] <= 10^9
1 <= bookings[i][2] <= 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 max_value(bookings: list[list[int]], k: int) -> int:

Java

public int maxValue(int[][] bookings, int k)

The intended approach

Time O(n log n + n * k) over the bookings and the allowance.Space O(n * k) for the walk.

Order the bookings by the day they finish. Then, walking them in that order, whether a booking can be added depends only on which bookings finish before it starts, and those form a run from the very beginning of the order. So one search per booking finds the cut-off: the first position whose finishing day is at or after this booking's first day. Everything before that cut-off is compatible. That turns the whole thing into a walk over two counters: how many bookings have been considered and how many have been taken. At each step the choice is to skip the booking, keeping whatever the earlier bookings managed with the same allowance, or to take it, adding its fee to whatever the bookings before its cut-off managed with one less allowance. Because the allowance is at most `k` and the two multiplied together are capped at a million, the walk fits. Days are inclusive at both ends, so a booking starting on the very day another finishes is a clash, which is why the cut-off is the first finishing day at or after the first day rather than after it.

Pythonentry point: max_value

def max_value(bookings, k):
    from bisect import bisect_left
    order = sorted(bookings, key=lambda b: b[1])
    n = len(order)
    ends = [b[1] for b in order]
    cut = [0] * n
    for i in range(n):
        cut[i] = bisect_left(ends, order[i][0])
    best = [[0] * (k + 1) for _ in range(n + 1)]
    for i in range(1, n + 1):
        for t in range(k + 1):
            skip = best[i - 1][t]
            take = 0
            if t > 0:
                take = best[cut[i - 1]][t - 1] + order[i - 1][2]
            best[i][t] = skip if skip > take else take
    return best[n][k]

The obvious version

Try every selection of bookings as a bit pattern, throw away the ones holding more than `k`, order each selection by its first day and walk it checking that every booking starts after the previous one finished, then keep the largest total fee. Costs two to the power of the booking count.

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_value(bookings, k):
    n = len(bookings)
    best = 0
    for mask in range(1 << n):
        picked = []
        for i in range(n):
            if mask >> i & 1:
                picked.append(bookings[i])
        if len(picked) > k:
            continue
        picked.sort(key=lambda b: b[0])
        ok = True
        total = 0
        reach = 0
        for b in picked:
            if b[0] <= reach:
                ok = False
                break
            total += b[2]
            reach = b[1]
        if ok and total > best:
            best = total
    return best

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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