1092MediumArrayQueueSliding WindowHeap (Priority Queue)Ordered SetMonotonic Queue

Stretches Whose Readings Stay Within Two

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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 log holds the readings readings. A stretch of neighbouring readings is steady when its largest reading and its smallest differ by at most 2. A stretch of one reading is always steady.

Return how many steady stretches the log holds.

Examples

Example 1

Input: readings = [1, 3, 5]
Output: 5

The three single readings are steady, and so are the two neighbouring pairs, each spread exactly two. The whole log spreads four, so it is not.

Example 2

Input: readings = [7, 7, 7]
Output: 6

Every reading is the same, so every stretch has no spread at all and all six count.

Example 3

Input: readings = [1, 10]
Output: 2

The two readings spread nine, so only the two single readings are steady.

Constraints

1 <= readings.length <= 10^5
1 <= readings[i] <= 10^9

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 continuous_subarrays(readings: list[int]) -> int:

Java

public long continuousSubarrays(int[] readings)

The intended approach

Time O(n) over the log's length.Space O(n) for the two lists of positions.

Widening a stretch never narrows its spread, so for a fixed right end the steady stretches are exactly those whose left end sits at or after some earliest place. That makes one window sliding along the log enough: for each right end, push the left end in until the window is steady again, then every left end from there to the right end gives a steady stretch, which is the window's length. The window needs its largest and smallest readings to hand as it changes at both ends. Keep two lists of positions, one holding readings that fall and one holding readings that rise, each trimmed from the back whenever a new reading makes an entry pointless: a reading no larger than the newcomer can never be the window's largest again. The fronts of the two lists then name the window's largest and smallest, and a position dropped from the window's left end is dropped from a list's front when it turns up there. Each position joins and leaves each list once, so the whole thing is one pass.

Pythonentry point: continuous_subarrays

def continuous_subarrays(readings):
    from collections import deque
    highs = deque()
    lows = deque()
    total = 0
    lo = 0
    for hi in range(len(readings)):
        value = readings[hi]
        while highs and readings[highs[len(highs) - 1]] <= value:
            highs.pop()
        highs.append(hi)
        while lows and readings[lows[len(lows) - 1]] >= value:
            lows.pop()
        lows.append(hi)
        while readings[highs[0]] - readings[lows[0]] > 2:
            if highs[0] == lo:
                highs.popleft()
            if lows[0] == lo:
                lows.popleft()
            lo += 1
        total += hi - lo + 1
    return total

The obvious version

Take every stretch outright: for each starting place walk forward keeping the largest and smallest reading seen, counting the stretch while the two stay within two of each other and stopping that start as soon as they do not, since widening only makes the spread worse. Costs the log's length squared rather than one window.

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_continuous_subarrays(readings):
    n = len(readings)
    total = 0
    for start in range(n):
        low = readings[start]
        high = readings[start]
        for end in range(start, n):
            if readings[end] < low:
                low = readings[end]
            if readings[end] > high:
                high = readings[end]
            if high - low > 2:
                break
            total += 1
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 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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