0338MediumArraySortingBucket SortRadix SortPigeonhole Principle

Widest Step Between Sorted Readings

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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 logger dumps nums, a batch of meter readings in the order they were captured, so the batch is in no particular order and readings may repeat.

Imagine the batch laid out on a number line from smallest to largest. Between each neighbouring pair of that laid-out sequence there is a step, and the step is the difference between the two values. Return the width of the widest step. When the batch holds fewer than two readings there is no step at all, so return 0.

Solve it in time that grows linearly with the number of readings, using extra space that also grows only linearly.

Examples

Example 1

Input: nums = [3, 41, 7, 20]
Output: 21

Laid out the readings run 3, 7, 20, 41, so the steps are 4, 13 and 21 and the widest is 21.

Example 2

Input: nums = [7, 7, 1]
Output: 6

Laid out the readings run 1, 7, 7, giving steps of 6 and 0, so the widest step is 6.

Example 3

Input: nums = [9]
Output: 0

A single reading leaves no neighbouring pair, so the answer is 0.

Example 4

Input: nums = [5, 5, 5]
Output: 0

All three readings match, so every step is 0.

Constraints

1 <= nums.length <= 10^5
0 <= nums[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 widest_reading_step(readings: list[int]) -> int:

Java

public int widestReadingStep(int[] readings)

The intended approach

Time O(n)Space O(n)

With n readings the laid-out sequence has n-1 steps that together span the whole range, so the widest step is at least the range divided by n-1. Bucketing the readings by that width means two readings sharing a bucket are closer than the widest step can be, so only the smallest reading of each occupied bucket paired against the largest of the previous occupied bucket needs measuring. One pass finds the extremes, one fills the buckets, one sweeps them.

Pythonentry point: widest_reading_step

def widest_reading_step(readings):
    # Pigeonhole: with n readings there are n-1 steps once sorted, so the widest
    # step is at least the average width. Slotting readings into buckets of that
    # width means no two readings in one bucket can be the widest step, and only
    # the low end of each occupied bucket against the high end of the previous
    # occupied bucket has to be measured.
    # Time O(n). Space O(n) for the bucket ends.
    n = len(readings)
    if n < 2:
        return 0
    lo = readings[0]
    hi = readings[0]
    for v in readings:
        if v < lo:
            lo = v
        if v > hi:
            hi = v
    if hi == lo:
        return 0
    width = -(-(hi - lo) // (n - 1))
    slots = (hi - lo) // width + 1
    bmin = [None] * slots
    bmax = [None] * slots
    for v in readings:
        k = (v - lo) // width
        if bmin[k] is None or v < bmin[k]:
            bmin[k] = v
        if bmax[k] is None or v > bmax[k]:
            bmax[k] = v
    best = 0
    prev = None
    for k in range(slots):
        if bmin[k] is None:
            continue
        if prev is not None and bmin[k] - prev > best:
            best = bmin[k] - prev
        prev = bmax[k]
    return best

The obvious version

Sort the batch outright and take the largest difference between neighbouring entries.

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_widest_reading_step(readings):
    if len(readings) < 2:
        return 0
    order = sorted(readings)
    best = 0
    for i in range(1, len(order)):
        d = order[i] - order[i - 1]
        if d > best:
            best = d
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
4 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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