1072MediumArrayMathSorting

Bringing Every Gauge to One Reading

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Trains the technique from

LeetCode 462Minimum Moves to Equal Array Elements II

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 rail of gauges reads gauges. One turn changes a single gauge by one, up or down.

Return the fewest turns that leave every gauge on the same reading.

Examples

Example 1

Input: gauges = [1, 1, 1]
Output: 0

Every gauge already reads the same, so nothing needs turning.

Example 2

Input: gauges = [1, 5]
Output: 4

The two readings are four apart, and any reading between them costs exactly four turns altogether.

Example 3

Input: gauges = [4, 4, 4, 4, 10]
Output: 6

Settling on four leaves the four gauges already there untouched and brings the ten down six turns. Settling anywhere higher would move four gauges instead of one.

Constraints

1 <= gauges.length <= 10^5
-10^9 <= gauges[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 min_moves2(gauges: list[int]) -> int:

Java

public long minMoves2(int[] gauges)

The intended approach

Time O(n log n) for the sort, then one pass.Space O(n) for the sorted readings.

Whatever reading they all end on, the cost is how far each gauge sits from it added up, so the question is which reading makes that total smallest. Pair the outermost two readings. Every reading between them costs exactly the gap between the pair, and anything outside costs more, so the pair is happy anywhere in between. Strip that pair away and repeat on what is left; the ranges only narrow and never conflict, so a middle reading suits every pair at once, and the smallest total is those outermost gaps taken pair by pair inward. So sort the readings and walk inward from both ends, adding gaps. Equivalently, settle on the middle reading of the sorted rail and add up the distances; both give the same number.

Pythonentry point: min_moves2

def min_moves2(gauges):
    order = sorted(gauges)
    total = 0
    lo = 0
    hi = len(order) - 1
    while lo < hi:
        total += order[hi] - order[lo]
        lo += 1
        hi -= 1
    return total

The obvious version

Try every reading that appears on the rail as the one to settle on, adding up how far each gauge sits from it, and keep the smallest total. Only readings on the rail need trying, since the best reading can always be shifted onto one of them without cost. Costs the rail's length squared rather than one inward walk.

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_moves2(gauges):
    best = None
    for target in gauges:
        total = 0
        for value in gauges:
            total += value - target if value > target else target - value
        if best is None or total < best:
            best = total
    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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