0368HardArrayQueueSliding WindowMonotonic Queue

Stretches Pinned to Both Limits

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Trains the technique from

LeetCode 2444Count Subarrays With Fixed Bounds

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 rig writes one pressure reading per second into readings.

A stretch is one or more readings that sit next to each other in the log. Two stretches count as different when they begin or end at different positions, even if the readings inside them happen to match.

A stretch is pinned when the smallest reading inside it is exactly low and the largest reading inside it is exactly high.

Return how many pinned stretches the log contains. Nothing guarantees that low is below high.

Examples

Example 1

Input: readings = [2,5,1,6,3,5], low = 1, high = 5
Output: 2

The reading 6 at position 3 is above high, so no pinned stretch may contain it. The two that qualify are positions 1 to 2, giving [5,1], and positions 0 to 2, giving [2,5,1]; each has 1 as its smallest reading and 5 as its largest.

Example 2

Input: readings = [4,4,4], low = 4, high = 4
Output: 6

Every stretch has 4 as both its smallest and its largest reading, and a log of three readings has six stretches.

Example 3

Input: readings = [3,3], low = 1, high = 3
Output: 0

No reading equals 1, so no stretch can have 1 as its smallest reading.

Example 4

Input: readings = [2,3], low = 3, high = 2
Output: 0

A stretch would need its smallest reading to be 3 and its largest to be 2, which nothing can satisfy.

Constraints

2 <= readings.length <= 10^5
1 <= readings[i], low, high <= 10^6

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_subarrays(readings: list[int], low: int, high: int) -> int:

Java

public long countSubarrays(int[] readings, int low, int high)

The intended approach

Time O(n)Space O(1)

Walk the log once and keep three positions: the last place `low` was read, the last place `high` was read, and the last place a reading fell outside the closed span from `low` to `high`. A stretch ending at the current position is pinned exactly when it starts after that outside reading and still reaches back over both of the other two positions, so the number of valid starts is the earlier of the two marker positions minus the outside position, whenever that is positive. Adding that at every position gives the total.

Pythonentry point: count_subarrays

def count_subarrays(readings, low, high):
    # One pass, three positions tracked. O(n) time, O(1) space.
    total = 0
    last_low = -1
    last_high = -1
    last_out = -1
    for i, value in enumerate(readings):
        if value < low or value > high:
            last_out = i
        if value == low:
            last_low = i
        if value == high:
            last_high = i
        reach = last_low if last_low < last_high else last_high
        if reach > last_out:
            total += reach - last_out
    return total

The obvious version

Fix each starting position and extend the end one reading at a time, keeping the running smallest and largest, and add one whenever the pair matches the two limits.

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_subarrays(readings, low, high):
    n = len(readings)
    total = 0
    for i in range(n):
        smallest = readings[i]
        largest = readings[i]
        for j in range(i, n):
            if readings[j] < smallest:
                smallest = readings[j]
            if readings[j] > largest:
                largest = readings[j]
            if smallest == low and largest == high:
                total += 1
    return total

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
4 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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