0194MediumArrayBinary Search

Wrapped Buoy Low Offset

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Trains the technique from

LeetCode 153Find Minimum in Rotated Sorted Array

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 weather buoy stores calibration offsets in a ring buffer. The offsets were written into the ring in strictly increasing order, one per slot, and afterwards the buffer's start marker was slid forward by some number of slots. Dumping the ring from the marker therefore hands you the increasing run cut at one slot, with the piece before the cut moved behind the piece after it. The marker may also have been slid all the way around the ring, and then the dump comes out plainly increasing.

Offsets can be below freezing, so negative values are ordinary. No two offsets are equal.

Given the dump offsets, report the smallest offset it holds. Your routine must settle the answer in logarithmic time, so reading every slot is off the table.

Examples

Example 1

Input: offsets = [-4, -1, 3, -9, -7]
Output: -9

The increasing run was -9, -7, -4, -1, 3; the cut fell after -1, so the run restarts at slot 3 and the low offset sits there.

Example 2

Input: offsets = [2, 7, 9]
Output: 2

The marker travelled the full ring, so the dump is still increasing and the low offset is the first slot.

Example 3

Input: offsets = [5000, -5000]
Output: -5000

Two slots with the cut between them, so the second slot restarts the run.

Constraints

n == offsets.length
1 <= n <= 5000
-5000 <= offsets[i] <= 5000
No two offsets are equal
offsets is a strictly increasing run whose start marker was slid forward between 1 and n slots

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

Java

public int lowestOffset(int[] offsets)

The intended approach

Time O(log n)Space O(1)

Hold a window that always contains the restart slot. Compare the middle slot against the window's last slot: if the middle reads higher, the cut lies somewhere to its right, so the restart cannot be at or left of the middle. Otherwise the middle is itself part of the run that ends the window, so the restart is at the middle or left of it. The window halves each round and closes on the low offset.

Pythonentry point: lowest_offset

def lowest_offset(offsets):
    # Time O(log n): the candidate window halves on every round.
    # Space O(1): two indices, nothing allocated.
    lo = 0
    hi = len(offsets) - 1
    while lo < hi:
        mid = (lo + hi) // 2
        if offsets[mid] > offsets[hi]:
            lo = mid + 1
        else:
            hi = mid
    return offsets[lo]

The obvious version

Read every slot and keep the running smallest, O(n).

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_lowest_offset(offsets):
    best = offsets[0]
    for value in offsets[1:]:
        if value < best:
            best = value
    return best

How this problem was checked

Reference solution executed against 12 test cases — 12 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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