0875EasyArraySimulation

Settling the Gauge Row

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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 row of gauges reads arr. Every round, all the gauges are updated at once from the readings the row held at the start of that round:

a gauge with a neighbour on each side rises by one when both neighbours read higher than it, and falls by one when both read lower;
otherwise it keeps its reading, and the gauges at the two ends never change.

Rounds continue until a round changes nothing. Return the row as it then stands.

Examples

Example 1

Input: arr = [1, 3, 2, 3]
Output: [1, 2, 3, 3]

In the first round the gauge reading 3 at position 1 has 1 and 2 either side, both lower, so it falls to 2, while the gauge reading 2 has 3 and 3 either side, both higher, so it rises to 3. The next round changes nothing.

Example 2

Input: arr = [5, 1, 5]
Output: [5, 5, 5]

The middle gauge rises by one each round while both neighbours stay higher, and it stops once it reaches them.

Example 3

Input: arr = [1, 2, 3]
Output: [1, 2, 3]

The middle gauge has one neighbour above and one below, so it never moves, and the two ends never move either.

Constraints

3 <= arr.length <= 100
1 <= arr[i] <= 100

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 settle_gauges(arr: list[int]) -> list[int]:

Java

public List<Integer> settleGauges(int[] arr)

The intended approach

Time O(n * r) for n gauges over r rounds until nothing changes.Space O(n) for the copy each round.

Because every gauge is updated from the readings at the start of the round, each round must be worked out into a fresh copy rather than in place, or a gauge would see a neighbour that has already moved. Compare each inner gauge against its two neighbours in the old copy, adjust it by one where both point the same way, and stop as soon as a round produces the copy it started from. Each round narrows the gaps, so the process always settles.

Pythonentry point: settle_gauges

def settle_gauges(arr):
    row = list(arr)
    n = len(row)
    while True:
        nxt = list(row)
        for i in range(1, n - 1):
            if row[i - 1] > row[i] and row[i + 1] > row[i]:
                nxt[i] = row[i] + 1
            elif row[i - 1] < row[i] and row[i + 1] < row[i]:
                nxt[i] = row[i] - 1
        if nxt == row:
            return row
        row = nxt

The obvious version

Run the same rounds but record every row the process has produced, and stop when a row turns up that has been seen before, then return it. Keeps the whole history instead of comparing against the previous row alone.

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_settle_gauges(arr):
    row = list(arr)
    n = len(row)
    seen = []
    while True:
        if row in seen:
            return row
        seen.append(list(row))
        nxt = list(row)
        for i in range(1, n - 1):
            left = row[i - 1]
            right = row[i + 1]
            if left > row[i] and right > row[i]:
                nxt[i] = row[i] + 1
            if left < row[i] and right < row[i]:
                nxt[i] = row[i] - 1
        row = nxt

How this problem was checked

Reference solution executed against 13 test cases — 13 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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