0661MediumArraySliding Window

Windows That Reach the Average Bar

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Trains the technique from

LeetCode 1343Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold

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 rooftop sensor writes one integer reading per minute into the list readings. An analyst inspects every window of exactly span consecutive readings and keeps the window when the mean of its readings is at least bar.

Two windows are different when they start at different minutes, so a list of n readings has n - span + 1 windows to inspect.

Return how many of those windows the analyst keeps.

Examples

Example 1

Input: readings = [7, 2, 9, 4, 6, 3, 8], span = 3, bar = 6
Output: 2

The five windows total 18, 15, 19, 13 and 17, so their means are 6, 5, 6.33..., 4.33... and 5.66.... Only the first and the third reach a mean of 6.

Example 2

Input: readings = [8, 1, 1, 8], span = 2, bar = 4
Output: 2

The three windows total 9, 2 and 9, giving means 4.5, 1 and 4.5. The first and the last reach the bar of 4.

Example 3

Input: readings = [4, 4, 4, 4, 4], span = 5, bar = 3
Output: 1

The span equals the number of readings, so there is one window; its mean is 4, which reaches the bar.

Constraints

1 <= readings.length <= 10^5
1 <= readings[i] <= 10^4
1 <= span <= readings.length
0 <= bar <= 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 count_qualifying_windows(readings: list[int], span: int, bar: int) -> int:

Java

public int countQualifyingWindows(int[] readings, int span, int bar)

The intended approach

Time O(n): the first window costs span additions and each later window costs two.Space O(1): a running total, the target and a counter.

A window's mean reaches `bar` exactly when its total reaches `span * bar`, so the whole problem can be settled with integer totals. Add up the first `span` readings, then move the window one minute at a time by adding the reading that entered and subtracting the one that left, testing the running total against `span * bar` at every position.

Pythonentry point: count_qualifying_windows

def count_qualifying_windows(readings, span, bar):
    # Comparing the window total against span * bar avoids any division.
    # O(n) time, O(1) space.
    need = span * bar
    total = 0
    for i in range(span):
        total += readings[i]
    good = 1 if total >= need else 0
    for right in range(span, len(readings)):
        total += readings[right] - readings[right - span]
        if total >= need:
            good += 1
    return good

The obvious version

For every start minute, add up that window's readings from scratch, divide by the span and compare the mean against the bar.

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

from fractions import Fraction


def bf_count_qualifying_windows(readings, span, bar):
    n = len(readings)
    good = 0
    for start in range(n - span + 1):
        acc = 0
        for i in range(start, start + span):
            acc += readings[i]
        if Fraction(acc, span) >= Fraction(bar):
            good += 1
    return good

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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