Fastest 1000 Meters: Max Average Speed from 5-Second Distance Splits

A runner's workout is recorded as a list of integers `distances`, where `distances[i]` is the number of meters covered during the `i`-th consecutive 5-second interval. Interval `i` covers the time from second `5 * i` to second `5 * i + 5`. Find the runner's fastest 1000 meters: the maximum average speed, in meters per second, over any continuous stretch of time during which the runner covered exactly 1000 meters. A stretch may begin and end part-way through an interval. The original assessment required linear interpolation in that case without spelling out the details; this version fixes them as follows: within an interval the runner moves at a constant speed, so after `t` seconds of interval `i` (with `0 <= t <= 5`) the runner has covered `distances[i] * t / 5` meters of that interval. ### Function Signature ```python def fastest_1000m_speed(distances: list[int]) -> float: ``` ### Rules - A valid stretch is a contiguous time span `[a, b]` with `0 <= a < b <= 5 * len(distances)` during which the runner covers exactly 1000 meters under the constant-speed-within-an-interval model. - The average speed of a stretch is `1000 / (b - a)` meters per second. Intervals with 0 meters that lie inside a stretch still add their time. - A stretch may lie entirely inside one interval when that interval alone covers at least 1000 meters. - Return the maximum average speed over all valid stretches. - The exact optimum is a single real number. A returned value is accepted if its relative error from the exact optimum is at most 1% (0.01), matching the tolerance of the original assessment. ### Constraints - `1 <= len(distances) <= 10^5` - `0 <= distances[i] <= 10^5` - `sum(distances) >= 1000`, so at least one valid stretch always exists. - `sum(distances)` can reach `10^10`, which exceeds `2^31 - 1`; use 64-bit or arbitrary-precision integers for running totals. All values stay well within `2^53`. ### Examples **Example 1** ```text Input: distances = [200, 200, 200, 200, 200] Output: 40.0 ``` The only stretch covering exactly 1000 meters is the whole 25-second run: `1000 / 25 = 40`. **Example 2** ```text Input: distances = [100, 400, 400, 300] Output: 75.0 ``` Intervals 1 and 2 cover 800 meters in 10 seconds. The remaining 200 meters come from the start of interval 3, where the runner moves at `300 / 5 = 60` meters per second, which takes `10/3` seconds. The stretch lasts `40/3` seconds, so the speed is `1000 / (40/3) = 75`. No other stretch is faster. **Example 3** ```text Input: distances = [1500] Output: 300.0 ``` The single interval is run at `1500 / 5 = 300` meters per second, so any 1000 meters inside it take `10/3` seconds: `1000 / (10/3) = 300`.