0975MediumMathBinary Search

Two Vans Making Every Drop

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Trains the technique from

LeetCode 3733Minimum Time to Complete All Deliveries

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.

Two vans make deliveries. Van i has d[i] drops to make and needs r[i] minutes for each drop it makes.

The two vans work at the same time as each other. Either van may take over any number of the other's drops, so the drops may be shared out between them however you like, but every drop must be made.

Return the fewest minutes until all the drops are done, which is the longer of the two vans' working times.

Examples

Example 1

Input: d = [14, 27], r = [5, 3]
Output: 78

There are 41 drops between them. In 63 minutes the first van manages 12 and the second 21, which is 33 and not enough; the smallest allowance that covers all 41 is what is returned.

Example 2

Input: d = [5, 5], r = [2, 2]
Output: 10

Ten drops between two vans of the same speed means five each, at two minutes apiece.

Example 3

Input: d = [10, 1], r = [2, 7]
Output: 18

Eleven drops in all. The fast van is worth more than three of the slow one's, so it takes most of them, and the two finish together as nearly as whole drops allow.

Constraints

d.length == 2
r.length == 2
1 <= d[i] <= 10^9
2 <= r[i] <= 3 * 10^4

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 minimum_time(d: list[int], r: list[int]) -> int:

Java

public long minimumTime(int[] d, int[] r)

The intended approach

Time O(log T) for a time range T: each candidate costs two divisions.Space O(1).

Only the total number of drops matters, since either van may make any of them, so the question is how to split that total between two vans of different speeds. Ask it the other way round: given a number of minutes, how many drops can be made? Each van manages its allowance divided by its own minutes per drop, rounded down, and the two together must cover the total. That test only becomes easier as the allowance grows, so the smallest allowance that works can be found by halving a range. The top of that range is the slower van doing everything alone, which is certainly enough.

Pythonentry point: minimum_time

def minimum_time(d, r):
    total = d[0] + d[1]
    slowest = r[0] if r[0] > r[1] else r[1]
    low = 0
    high = slowest * total
    while low < high:
        mid = (low + high) // 2
        made = mid // r[0] + mid // r[1]
        if made >= total:
            high = mid
        else:
            low = mid + 1
    return low

The obvious version

Try every way of splitting the drops between the two vans, work out each van's time and take the longer, then keep the smallest such figure. Costs a step per possible split, so it only works at small totals, but it makes no argument about allowances.

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_minimum_time(d, r):
    total = d[0] + d[1]
    best = None
    for first in range(total + 1):
        second = total - first
        a = first * r[0]
        b = second * r[1]
        longer = a if a > b else b
        if best is None or longer < best:
            best = longer
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 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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