0404MediumArrayMath

Calibration Pulses To Level The Gauges

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Trains the technique from

LeetCode 453Minimum Moves to Equal Array Elements

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 rack carries n pressure gauges, and offsets[i] is the signed zero-offset of gauge i measured in ticks. An offset can sit below zero.

A pulse singles out one gauge and lifts the offset of every gauge except that one by exactly one tick. The singled-out gauge is untouched.

Return the fewest pulses that leave all n offsets reading the same value. The answer always fits in a signed 32-bit integer.

Examples

Example 1

Input: offsets = [4,7,9]
Output: 8

Eight pulses can settle the rack at 12: gauge 1 is the singled-out gauge on 3 of them, gauge 2 on the other 5, and gauge 0 on none. Gauge 0 is lifted all 8 times to 12, gauge 1 is lifted 5 times to 12, and gauge 2 is lifted 3 times to 12.

Example 2

Input: offsets = [-3,-3]
Output: 0

Both offsets already read the same value, so no pulse is needed.

Example 3

Input: offsets = [-5,0,2,2]
Output: 19

Nineteen pulses can settle the rack at 14, with gauge 0 singled out on none of them, gauge 1 on 5 of them, and gauges 2 and 3 on 7 each. Offsets may start below zero, as gauge 0 does here.

Example 4

Input: offsets = [6]
Output: 0

A rack of one gauge is already uniform, so the count is zero.

Example 5

Input: offsets = [8,3]
Output: 5

Five pulses can settle the rack at 8, with gauge 0 singled out on every one of them, so only gauge 1 is ever lifted.

Constraints

n == offsets.length
1 <= n <= 10^5
-10^9 <= offsets[i] <= 10^9
The answer is guaranteed to fit in a signed 32-bit integer.

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_moves(offsets: list[int]) -> int:

Java

public int minMoves(int[] offsets)

The intended approach

Time O(n)Space O(1)

Lifting every gauge but one by a tick shifts the whole rack the same way as dropping that one gauge by a tick, so the pulse count equals the total distance each gauge has to fall to reach the lowest offset in the rack. A single pass accumulates the sum of the offsets and tracks the smallest one, and the answer is the sum less n copies of that smallest offset.

Pythonentry point: min_moves

def min_moves(offsets):
    # Time O(n): a single pass that carries the running total and the smallest
    # offset together. Space O(1) - two scalars, nothing built.
    smallest = offsets[0]
    total = 0
    for value in offsets:
        total += value
        if value < smallest:
            smallest = value
    return total - smallest * len(offsets)

The obvious version

Sort the offsets, then walk the sorted copy adding up how far each entry sits above the first one.

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_moves(offsets):
    ordered = sorted(offsets)
    floor = ordered[0]
    pulses = 0
    for value in ordered:
        pulses += value - floor
    return pulses

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
5 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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