Date: 2026-06-06
Time: 17:52
This file solves LeetCode 2335: Minimum Amount of Time to Fill Cups. A water dispenser can fill at most 2 cups of different types per second (or 1 cup of any type). Given three cup counts, find the minimum seconds to fill all of them.
min_seconds(amount: list[int]) -> int — The sole function. It computes the answer as a closed-form expression rather than simulating the process:
return max(max(amount), (sum(amount) + 1) // 2)
This takes the maximum of two lower bounds:
1. max(amount) — The dominant-type bound. Even if you pair the largest type with a different type every second, you still need at least this many seconds.
2. (sum(amount) + 1) // 2 — The throughput bound. You fill at most 2 cups per second, so you need at least ceil(total / 2) seconds. The +1 with integer division implements ceiling without floating point.
Closed-form over simulation. Many greedy solutions to this problem sort and re-pick the two largest counts each round (O(n * max) with a heap). This solution recognizes that both lower bounds are always *achievable* — you can always construct a valid filling schedule that hits whichever bound is tighter — so it collapses to O(1).
Ceiling division idiom. (n + d - 1) // d is the standard Python integer ceiling division pattern. Here d = 2, so it's (sum + 1) // 2.
minimum-amount-of-time-to-fill-cups/test_solution.py (directly). The large "Imported By" list in the prompt is an artifact of the repo's test harness structure — those other test files don't actually import this solution.1. Compute max(amount) — the single largest cup count.
2. Compute (sum(amount) + 1) // 2 — ceiling of half the total.
3. Return whichever is larger.
No loops, no branching, no mutation.
amount must contain exactly 3 non-negative integers (per the problem constraints). The function doesn't validate this — it relies on the caller (LeetCode harness) to guarantee it.>= max(amount) and >= ceil(sum(amount) / 2).None. An empty list would raise on max(amount). Negative values would produce a mathematically valid but semantically meaningless result. Both are outside the problem's constraint space.
The key insight is a proof sketch: if max(amount) >= sumofothertwo, the largest type dominates and you pair it with the others until they run out, then fill the rest alone — exactly max(amount) seconds. If max(amount) < sumofothertwo, you can always interleave the three types to fill 2 cups every second (except possibly the last), giving ceil(total / 2). In neither case can you do better, so these bounds are tight.
minimum-amount-of-time-to-fill-cups/test_solution.py — See which edge cases are tested (all zeros, one dominant type, equal distribution)minimum-amount-of-time-to-fill-cups/plan.md — The planning doc may capture the reasoning path from simulation to closed-formgreedy-to-closed-form-reductions — Other problems in this repo where a greedy simulation collapses to O(1) math (e.g., count-of-matches-in-tournament, minimum-cuts-to-divide-a-circle)minimum-amount-of-time-to-fill-cups/review.md — Review notes may discuss the correctness proof or alternative approaches consideredfill-cups-o1-time — min_seconds runs in O(1) time and space regardless of input valuesfill-cups-two-lower-bounds — The result is always max(max(amount), ceil(sum(amount)/2)), and both bounds are achievablefill-cups-no-validation — The function assumes exactly 3 non-negative integers; it performs no input validationfill-cups-ceiling-div-idiom — (sum(amount) + 1) // 2 computes ceil(sum / 2) using integer arithmetic, avoiding float precision issues