0218EasyArrayStackMonotonic Stack

Job Hour Credits

Tracked in this browser only

Trains the technique from

LeetCode 1475Final Prices With a Special Discount in a Shop

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 keeps its jobs in one queue and bills them in queue order. hours holds the hours booked against each job, position by position.

One settlement rule applies to every job. Read further along the queue from that job and stop at the first job whose booked hours are no more than this job's booked hours; that many hours come off this job's bill. If no later job qualifies, the job is billed exactly as booked.

The rule is worked out for each job from the booked figures alone, so a credit given to one job never changes what any other job is credited, and a job is never credited against itself.

Return the bill for every job, in queue order.

Examples

Example 1

Input: hours = [9, 5, 5, 12, 4]
Output: [4, 0, 1, 8, 4]

Job 0 stops at the 5 in position 1, so it is billed 9 - 5. Job 1 stops at the equal 5 in position 2 and is billed 5 - 5. Job 2 stops at the 4 in position 4 and is billed 5 - 4. Job 3 stops at that same 4 and is billed 12 - 4. Job 4 has nothing after it, so its 4 hours stand.

Example 2

Input: hours = [7, 8, 10]
Output: [7, 8, 10]

Every job further along the queue is booked for more hours than the one before it, so no job finds a qualifying job and each is billed as booked.

Example 3

Input: hours = [12, 3, 12, 3]
Output: [9, 0, 9, 3]

Both 12-hour jobs stop at a 3 that follows them and are billed 9. The 3 in position 1 stops at the equal 3 in position 3 and is billed 0, while the last job keeps its 3 hours.

Constraints

1 <= hours.length <= 500
1 <= hours[i] <= 1000

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 job_hour_credits(hours: list[int]) -> list[int]:

Java

public int[] jobHourCredits(int[] hours)

The intended approach

Time O(n)Space O(n)

Start every bill at the hours booked, then walk the queue once holding the positions still waiting for a credit. Their booked hours run high to low from the bottom of that hold, so when the current job arrives it settles every waiting position booked for at least as many hours: each of those has just met its first qualifying job. Positions still waiting when the walk ends never found one and keep their booked figure.

Pythonentry point: job_hour_credits

def job_hour_credits(hours):
    bills = list(hours)
    waiting = []
    for i in range(len(hours)):
        h = hours[i]
        while waiting and hours[waiting[-1]] >= h:
            bills[waiting.pop()] -= h
        waiting.append(i)
    return bills

The obvious version

For each job scan forward from the next position until a job booked for no more hours turns up, O(n^2).

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_job_hour_credits(hours):
    n = len(hours)
    out = []
    for i in range(n):
        credit = 0
        for j in range(i + 1, n):
            if hours[j] <= hours[i]:
                credit = hours[j]
                break
        out.append(hours[i] - credit)
    return out

How this problem was checked

Reference solution executed against 13 test cases — 13 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