0748MediumArrayMathSortingMatrix

Levelling A Bench Of Stepped Dials

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Trains the technique from

LeetCode 2033Minimum Operations to Make a Uni-Value Grid

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 test bench carries a rectangular block of dials. grid[i][j] is the setting the dial in row i, column j currently shows.

One adjustment picks a single dial and either raises its setting by exactly step or lowers it by exactly step. Nothing else can change a dial. A setting is free to pass below zero or above its starting value along the way.

The bench is level when every dial shows the same setting. Return the fewest adjustments that leave the bench level, or -1 when no sequence of adjustments can level it.

Examples

Example 1

Input: grid = [[3, 7, 11], [15, 19, 23]], step = 4
Output: 9

Bring every dial to 15. The 3 takes three raises, the 7 takes two, the 11 takes one, the 15 takes none, the 19 takes one lowering and the 23 takes two, which is 3 + 2 + 1 + 0 + 1 + 2 = 9 adjustments.

Example 2

Input: grid = [[6, 7]], step = 3
Output: -1

From 6 the reachable settings are `..., 0, 3, 6, 9, 12, ...` and from 7 they are `..., 1, 4, 7, 10, 13, ...`. The two dials share no reachable setting, so the bench cannot be levelled.

Example 3

Input: grid = [[2, 2], [2, 8]], step = 2
Output: 3

Three lowerings of 2 turn the 8 into a 2, and the other three dials already show 2.

Constraints

1 <= grid.length <= 10^5
1 <= grid[i].length <= 10^5
1 <= grid.length * grid[i].length <= 10^5, so the bench holds at most 10^5 dials.
1 <= step <= 10^4
1 <= grid[i][j] <= 10^4
Every row of grid has the same 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 min_adjustments(grid: list[list[int]], step: int) -> int:

Java

public int minAdjustments(int[][] grid, int step)

The intended approach

Time O(N log N) for N dials, dominated by the sort.Space O(N) for the flattened list.

Flatten the bench into one list of settings, since the rows and columns carry no meaning here. An adjustment shifts a dial by exactly `step`, so the settings a dial can reach are exactly those leaving the same remainder as its current setting when divided by `step`; if two dials disagree on that remainder the bench can never be levelled, and the answer is `-1`. Otherwise sort the settings and drive every dial to the setting sitting at the middle of the sorted list, which balances the number of dials that must rise against the number that must fall. Each dial then costs its gap to that setting divided by `step`, and the gaps are all exact multiples because the remainders agree.

Pythonentry point: min_adjustments

def min_adjustments(grid, step):
    # every adjustment shifts a dial by exactly `step`, so a dial can only ever
    # reach settings that leave the same remainder as it does now.
    # O(N log N) time for N dials because of the sort, O(N) space.
    flat = []
    for row in grid:
        flat.extend(row)
    mark = flat[0] % step
    for v in flat:
        if v % step != mark:
            return -1
    flat.sort()
    target = flat[len(flat) // 2]
    total = 0
    for v in flat:
        gap = v - target
        if gap < 0:
            gap = -gap
        total += gap // step
    return total

The obvious version

After the remainder check, walk every setting reachable from the lowest dial in steps of `step` up to the highest dial, total the adjustments needed to bring all dials to that setting, and keep the smallest total. No balancing argument at all.

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_min_adjustments(grid, step):
    """Try every setting the dials could all be driven to, which are the settings
    reachable from the lowest dial without passing the highest, and total the
    adjustments for each."""
    flat = [v for row in grid for v in row]
    mark = flat[0] % step
    for v in flat:
        if v % step != mark:
            return -1
    lo = min(flat)
    hi = max(flat)
    best = None
    target = lo
    while target <= hi:
        total = 0
        for v in flat:
            total += abs(v - target) // step
        if best is None or total < best:
            best = total
        target += step
    return best

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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