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Trains the technique from

LeetCode 3635Earliest Finish Time for Land and Water Rides 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 trainee at a maker space has to sit exactly one lathe session and exactly one welding session, and cannot be at both benches at the same time.

Lathe session i opens at minute latheOpen[i] and, once begun, keeps the trainee busy for latheRuns[i] minutes. Welding session j opens at minute weldOpen[j] and keeps the trainee busy for weldRuns[j] minutes. A session may begin at any minute at or after the minute it opens, never before, and the trainee is free to wait around for as long as needed between sessions.

The trainee books one session at each bench and sits them one after the other, in whichever order suits. Return the earliest minute by which both sessions can be finished.

Examples

Example 1

Input: latheOpen = [3, 20], latheRuns = [40, 5], weldOpen = [1], weldRuns = [6]
Output: 25

Welding runs from minute 1 to minute 7, then the second lathe session opens at minute 20 and runs to minute 25, so both are done by minute 25.

Example 2

Input: latheOpen = [1], latheRuns = [10], weldOpen = [1], weldRuns = [10]
Output: 21

There is one choice at each bench and they cannot overlap, so one runs from minute 1 to minute 11 and the other from minute 11 to minute 21.

Example 3

Input: latheOpen = [2, 9], latheRuns = [3, 1], weldOpen = [14, 4], weldRuns = [2, 7]
Output: 12

The second lathe session runs from minute 9 to minute 10, and the welding session that opened at minute 14 runs to minute 16, so both finish by minute 16.

Constraints

1 <= latheOpen.length == latheRuns.length <= 10^5
1 <= weldOpen.length == weldRuns.length <= 10^5
1 <= latheOpen[i], weldOpen[j] <= 10^8
1 <= latheRuns[i], weldRuns[j] <= 10^8

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 earliest_both_done(latheOpen: list[int], latheRuns: list[int], weldOpen: list[int], weldRuns: list[int]) -> int:

Java

public int earliestBothDone(int[] latheOpen, int[] latheRuns, int[] weldOpen, int[] weldRuns)

The intended approach

Time O(n + m)Space O(1)

Fix which bench comes first. The only thing the first session hands to the second is the minute the trainee becomes free, and a smaller value there can never hurt, so the first bench collapses to the smallest value of open plus run across its sessions. With that single minute in hand, scan the other bench and take the smallest of max(open, free) + run. Do this for both orders and keep the smaller result.

Pythonentry point: earliest_both_done

def earliest_both_done(latheOpen, latheRuns, weldOpen, weldRuns):
    # Whichever bench is used first, only its earliest possible finishing minute
    # matters: finishing the first session later can never let the second end
    # sooner. So collapse the first bench to one number, then scan the second.
    # time O(n + m), space O(1)
    def pass_order(firstOpen, firstRun, secondOpen, secondRun):
        ready = min(o + r for o, r in zip(firstOpen, firstRun))
        best = None
        for o, r in zip(secondOpen, secondRun):
            done = (o if o > ready else ready) + r
            if best is None or done < best:
                best = done
        return best
    return min(pass_order(latheOpen, latheRuns, weldOpen, weldRuns),
               pass_order(weldOpen, weldRuns, latheOpen, latheRuns))

The obvious version

Try every lathe session against every welding session, and for each such pair work out both running orders explicitly, keeping the smallest finishing minute seen. Quadratic in the two bench sizes.

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_earliest_both_done(latheOpen, latheRuns, weldOpen, weldRuns):
    best = None
    for i in range(len(latheOpen)):
        for j in range(len(weldOpen)):
            a = latheOpen[i] + latheRuns[i]
            first = max(weldOpen[j], a) + weldRuns[j]
            b = weldOpen[j] + weldRuns[j]
            second = max(latheOpen[i], b) + latheRuns[i]
            for cand in (first, second):
                if best is None or cand < best:
                    best = cand
    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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