0276HardArrayDynamic ProgrammingGraph TheoryTopological SortDirected Acyclic Graph

Refinery Shutdown Span

Tracked in this browser only

Trains the technique from

LeetCode 2050Parallel Courses III

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 refinery shutdown is broken into stages jobs numbered 1 through stages. Job i occupies hours[i - 1] hours from the moment it starts and runs straight through to its finish.

precedes[j] = [a, b] records that job a has to be finished before job b may begin. Each ordered pair appears at most once and the pairs never close a loop.

The crew is large enough that any number of jobs may run side by side, and a job begins at the very moment the last job it waits on has finished. A job with nothing ahead of it begins at hour 0.

Return the number of hours from hour 0 until the final job finishes.

Examples

Example 1

Input: stages = 4, precedes = [[1, 2], [1, 3], [2, 4], [3, 4]], hours = [2, 3, 6, 1]
Output: 9

Job 1 runs from hour 0 to hour 2. Jobs 2 and 3 both begin at hour 2 and finish at hours 5 and 8. Job 4 waits on both of them, so it begins at hour 8 and finishes at hour 9.

Example 2

Input: stages = 5, precedes = [], hours = [3, 1, 4, 1, 5]
Output: 5

Nothing waits on anything, so all five jobs begin at hour 0 and job 5 is the last to finish, at hour 5.

Example 3

Input: stages = 3, precedes = [[1, 2]], hours = [5, 1, 9]
Output: 9

Job 1 runs from hour 0 to hour 5 and job 2 follows from hour 5 to hour 6. Job 3 waits on nothing, so it runs from hour 0 to hour 9.

Constraints

1 <= stages <= 5 * 10^4
0 <= precedes.length <= min(stages * (stages - 1) / 2, 5 * 10^4)
precedes[j].length == 2
1 <= a, b <= stages
a != b
All the pairs [a, b] are unique.
hours.length == stages
1 <= hours[i] <= 10^4
The pairs never close a loop.

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 shutdown_span(stages: int, precedes: list[list[int]], hours: list[int]) -> int:

Java

public int shutdownSpan(int stages, int[][] precedes, int[] hours)

The intended approach

Time O(stages + len(precedes))Space O(stages + len(precedes))

Build the forward adjacency and a count of unfinished predecessors per job, then settle jobs in an order that only reaches a job once every predecessor is settled. A ready list with a moving head index serves as the queue, so nothing recurses. Each settled job pushes its finish hour into its successors' start times, keeping the latest of them, and decrements their pending counts. The answer is the largest finish hour seen over all jobs, not the finish of whichever job was settled last.

Pythonentry point: shutdown_span

def shutdown_span(stages, precedes, hours):
    # Settle the jobs in an order where every job is reached only after all of its
    # predecessors, using a plain list as the ready queue with a moving head index
    # rather than recursion, so a chain covering all 50000 jobs is safe.
    # start[w] collects the latest finish among w's predecessors while w waits.
    # Time: O(stages + len(precedes)).  Space: O(stages + len(precedes)).
    after = [[] for _ in range(stages + 1)]
    waiting = [0] * (stages + 1)
    for a, b in precedes:
        after[a].append(b)
        waiting[b] += 1
    ready = [v for v in range(1, stages + 1) if waiting[v] == 0]
    start = [0] * (stages + 1)
    span = 0
    head = 0
    while head < len(ready):
        v = ready[head]
        head += 1
        done = start[v] + hours[v - 1]
        if done > span:
            span = done
        for w in after[v]:
            if done > start[w]:
                start[w] = done
            waiting[w] -= 1
            if waiting[w] == 0:
                ready.append(w)
    return span

The obvious version

Start each job's finish at its own duration, then sweep the whole pair list over and over pushing finish[a] + hours[b] into finish[b] whenever that is larger, until a full sweep changes nothing. Report the largest finish.

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_shutdown_span(stages, precedes, hours):
    finish = [0] * (stages + 1)
    for v in range(1, stages + 1):
        finish[v] = hours[v - 1]
    for _ in range(stages):
        changed = False
        for a, b in precedes:
            cand = finish[a] + hours[b - 1]
            if cand > finish[b]:
                finish[b] = cand
                changed = True
        if not changed:
            break
    best = 0
    for v in range(1, stages + 1):
        if finish[v] > best:
            best = finish[v]
    return best

How this problem was checked

Reference solution executed against 7 test cases — 7 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in