0285MediumArrayBinary Search

Reading Present in the Rotated Log

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Trains the technique from

LeetCode 81Search in Rotated Sorted Array 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 weather station files temperatures into a ring of fixed size. It files them in non-decreasing order and carries on around the ring, so a reading may equal the one filed before it.

A download starts at whichever slot the reader happens to open at and then walks the ring all the way round. What comes out is log: the non-decreasing run of temperatures shifted round by some number of slots. The shift may be zero, in which case log is simply the run itself.

Return true when the temperature wanted sits in some slot of log, and false otherwise.

Examples

Example 1

Input: log = [8, 9, 12, -3, -1, 3], wanted = -1
Output: true

The download opened three slots along, so the run 8, 9, 12 comes out first and -3, -1, 3 follows. The temperature -1 sits at index 4.

Example 2

Input: log = [4, -1, 4, 4, 4], wanted = -1
Output: true

The temperature -1 sits at index 1 of the download.

Example 3

Input: log = [8, 9, 12, -3, -1, 3], wanted = 7
Output: false

No slot of the download holds 7.

Constraints

1 <= log.length <= 5000
-10^4 <= log[i] <= 10^4
log is guaranteed to be shifted round at some slot.
-10^4 <= wanted <= 10^4

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 reading_present(log: list[int], wanted: int) -> bool:

Java

public boolean readingPresent(int[] log, int wanted)

The intended approach

Time O(log n) with distinct readings, O(n) worst case when readings repeatSpace O(1)

Hold a window and look at its middle slot. If the readings at the window's low end and its middle are in order then that half is an unshifted stretch, and the target is followed into it only when it lies between those two readings; otherwise the other half is the ordered one and the same test applies there. When the readings at the low end, the middle and the high end are all equal, neither half can be identified, so both ends move inward by a single slot, which loses nothing because both of those readings equal the middle one already checked.

Pythonentry point: reading_present

def reading_present(log, wanted):
    # Halve the window as in a plain bisection, but first work out which side of the
    # middle is still in order and only follow that side when the target lies inside
    # its span.  When the readings at the low end, the middle and the high end are
    # all equal, neither side can be told apart, so both ends step inward by one -
    # that is the case that drags the worst case out to a linear number of steps.
    # Time: O(log n) with distinct readings, O(n) in the worst case with repeats.
    # Space: O(1).
    lo = 0
    hi = len(log) - 1
    while lo <= hi:
        mid = (lo + hi) // 2
        if log[mid] == wanted:
            return True
        if log[lo] == log[mid] and log[mid] == log[hi]:
            lo += 1
            hi -= 1
        elif log[lo] <= log[mid]:
            if log[lo] <= wanted < log[mid]:
                hi = mid - 1
            else:
                lo = mid + 1
        else:
            if log[mid] < wanted <= log[hi]:
                lo = mid + 1
            else:
                hi = mid - 1
    return False

The obvious version

Walk the download from the first slot to the last and report success on the first slot that holds the wanted temperature.

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_reading_present(log, wanted):
    for v in log:
        if v == wanted:
            return True
    return False

How this problem was checked

Reference solution executed against 10 test cases — 10 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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