0321MediumArrayHash TableSliding WindowPrefix Sum

Stretches With Exact Press Count

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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 foot pedal on a press writes one entry per minute into a log. minutes[i] is 1 if the pedal was pressed during minute i and 0 if it was not.

A stretch is any run of one or more consecutive minutes from the log. Count the stretches that contain exactly goal presses. Two stretches are different when they start or end at different minutes, even when they hold the same entries.

Return that count.

Examples

Example 1

Input: minutes = [1, 1, 0, 1], goal = 2
Output: 3

Writing a stretch as the minutes it covers, the qualifying ones are minutes 0-1, minutes 0-2 and minutes 1-3. Each of those holds two presses.

Example 2

Input: minutes = [0, 1, 0, 0, 1], goal = 1
Output: 9

Six stretches hold only the press at minute 1: they start at minute 0 or 1 and end at minute 1, 2 or 3. Three more hold only the press at minute 4: they start at minute 2, 3 or 4 and end at minute 4.

Example 3

Input: minutes = [0, 0, 0], goal = 0
Output: 6

The log has no presses at all, so every stretch qualifies: three of length one, two of length two and one of length three.

Example 4

Input: minutes = [1, 0, 1, 1], goal = 0
Output: 1

The only stretch with no press is the single minute 1.

Example 5

Input: minutes = [1], goal = 0
Output: 0

The one available stretch holds a press, so nothing qualifies.

Constraints

1 <= minutes.length <= 3 * 10^4
minutes[i] is either 0 or 1.
0 <= goal <= minutes.length

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_stretches(minutes: list[int], goal: int) -> int:

Java

public int countStretches(int[] minutes, int goal)

The intended approach

Time O(n)Space O(n) for the tally of prefix counts

A stretch ending at minute `i` holds `running - earlier` presses, where `running` is the press count over minutes `0..i` and `earlier` is the count over the minutes cut off the front. So for each `i` the qualifying stretches are exactly the earlier prefixes whose count equals `running - goal`. Walk the log once keeping a tally of how often each prefix count has been seen, seeded with one empty prefix of count zero, and add the tally for `running - goal` at every step.

Pythonentry point: count_stretches

def count_stretches(minutes, goal):
    # O(n) time, one pass. O(n) space for the prefix-count tally.
    seen = [0] * (len(minutes) + 2)
    seen[0] = 1
    running = 0
    total = 0
    for v in minutes:
        running += v
        if running - goal >= 0:
            total += seen[running - goal]
        seen[running] += 1
    return total

The obvious version

Fix every start minute and extend the stretch one minute at a time, accumulating the press count and counting each time it lands on the goal.

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_count_stretches(minutes, goal):
    n = len(minutes)
    hits = 0
    for start in range(n):
        presses = 0
        for end in range(start, n):
            presses += minutes[end]
            if presses == goal:
                hits += 1
    return hits

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
5 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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