0320HardArrayGreedySortingHeap (Priority Queue)

Workshop Job Selection

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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 repair workshop has a board of n jobs it could take on and enough staff to finish at most max_jobs of them. Job i pays payouts[i] once it is delivered, but the shop will only be handed the job if it already has at least setup[i] in the bank, because that is the value of loan equipment the job needs.

The shop starts with funds in the bank. Taking a job does not spend anything: the setup figure is only a threshold the bank balance has to clear, and the payout is added to the balance as soon as the job is finished. Jobs are worked one after another, so a payout is available before choosing the next job, and each job can be taken at most once.

Choose at most max_jobs jobs to end up with as much money in the bank as possible, and return that final balance. The shop may also stop early, or take no jobs at all, if nothing on the board is within reach.

Examples

Example 1

Input: max_jobs = 2, funds = 3, payouts = [5, 8, 2], setup = [0, 6, 3]
Output: 16

With 3 in the bank, jobs 0 and 2 are within reach. Taking job 0 pays 5 and brings the balance to 8, which now clears job 1's threshold of 6. Taking job 1 pays 8 for a final balance of 16, and that is two jobs.

Example 2

Input: max_jobs = 1, funds = 0, payouts = [4, 9], setup = [0, 1]
Output: 4

An empty bank only clears job 0's threshold of 0, so the single job taken pays 4.

Example 3

Input: max_jobs = 4, funds = 2, payouts = [3, 1, 7], setup = [2, 0, 12]
Output: 6

Job 0 pays 3, taking the balance to 5, and job 1 pays 1, taking it to 6. Job 2 needs 12 in the bank, so the shop stops after two of its four allowed jobs.

Example 4

Input: max_jobs = 3, funds = 0, payouts = [6, 5], setup = [1, 2]
Output: 0

Both thresholds are above the starting balance of 0, so no job can be handed over and the balance stays at 0.

Constraints

1 <= max_jobs <= 10^5
0 <= funds <= 10^9
n == payouts.length
n == setup.length
1 <= n <= 10^5
0 <= payouts[i] <= 10^4
0 <= setup[i] <= 10^9

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_funds(max_jobs: int, funds: int, payouts: list[int], setup: list[int]) -> int:

Java

public int maxFunds(int maxJobs, int funds, int[] payouts, int[] setup)

The intended approach

Time O(n log n + k log n)Space O(n)

The bank balance never falls, so once a job is within reach it stays within reach. Order the jobs by their setup threshold and keep a pointer into that order: before each pick, advance the pointer over every job the current balance now clears and drop its payout into a max-heap. Then pop the largest payout available and add it to the balance. Repeat until `max_jobs` jobs are taken or the heap is empty.

Pythonentry point: max_funds

import heapq


def max_funds(max_jobs, funds, payouts, setup):
    # O(n log n) to order the jobs by setup requirement, then each of the at most
    # max_jobs rounds does heap work, giving O(n log n + k log n) overall.
    # O(n) space for the ordering and the heap.
    n = len(payouts)
    order = sorted(range(n), key=lambda i: setup[i])
    ready = []
    idx = 0
    for _ in range(max_jobs):
        while idx < n and setup[order[idx]] <= funds:
            heapq.heappush(ready, -payouts[order[idx]])
            idx += 1
        if not ready:
            break
        funds += -heapq.heappop(ready)
    return funds

The obvious version

Run at most max_jobs rounds and, in each round, scan the whole board again for the untaken job with the largest payout the current balance can clear.

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_funds(max_jobs, funds, payouts, setup):
    taken = [False] * len(payouts)
    for _ in range(max_jobs):
        pick = -1
        for i in range(len(payouts)):
            if taken[i] or setup[i] > funds:
                continue
            if pick == -1 or payouts[i] > payouts[pick]:
                pick = i
        if pick == -1:
            break
        taken[pick] = True
        funds += payouts[pick]
    return funds

How this problem was checked

Reference solution executed against 12 test cases — 12 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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