0111EasyArraySliding Window

Warmest Run of Days

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Trains the technique from

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 weather station keeps one figure per day for a site: the daily anomaly, how far that day sat above or below the long-run norm, measured in tenths of a degree. A cold day carries a negative figure.

You are given the array anomalies, one figure per day in date order, and an integer span. Look at every block of span days that run back to back, work out the mean anomaly of each block as the block total divided by span, and report the largest mean any block reaches.

Return the mean as a real number; a reported figure off by less than 1e-5 from the true one is accepted. Since span never exceeds the number of days on record, at least one block always exists.

Examples

Example 1

Input: anomalies = [-3, 14, 2, -8, 21, 6], span = 3
Output: 6.333333333333333

The four blocks total 13, 8, 15 and 19. The last three days total 19, the largest of them, so the mean reported is 19 divided by 3.

Example 2

Input: anomalies = [-9, -4], span = 2
Output: -6.5

Only one block of two days exists and both days ran cold, so the reported mean is negative.

Example 3

Input: anomalies = [7], span = 1
Output: 7.0

A single day forms the only block, and a block of one day has that day's figure as its mean.

Constraints

n == anomalies.length
1 <= span <= n <= 10^5
-10^4 <= anomalies[i] <= 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 warmest_run_average(anomalies: list[int], span: int) -> float:

Java

public double warmestRunAverage(int[] anomalies, int span)

The intended approach

Time O(n)Space O(1)

Every block holds the same number of days, so comparing means is the same as comparing block totals and the division can wait until the end. Total the first block, then slide one day at a time by adding the day that enters and subtracting the day that leaves, tracking the largest total seen. Dividing that single largest total by the span at the very end keeps the arithmetic in whole numbers throughout.

Pythonentry point: warmest_run_average

def warmest_run_average(anomalies, span):
    running = 0
    for i in range(span):
        running += anomalies[i]
    best = running
    for i in range(span, len(anomalies)):
        running += anomalies[i] - anomalies[i - span]
        if running > best:
            best = running
    return best / span

The obvious version

Walk every possible starting day and add up that block's figures from scratch, keeping the largest total, then divide once by the span. O(n * span).

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_warmest_run_average(anomalies, span):
    best = None
    for start in range(len(anomalies) - span + 1):
        total = 0
        for i in range(start, start + span):
            total += anomalies[i]
        if best is None or total > best:
            best = total
    return best / span

How this problem was checked

Reference solution executed against 12 test cases — 12 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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