0070MediumArrayBinary SearchMatrix

Paged Offset Lookup

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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 calibration archive keeps signed offset readings in fixed-width pages. You are given pages, a list of m pages of n readings each, laid out so that:

readings grow strictly from left to right inside a page, and
every reading on a page is strictly smaller than every reading on the page after it.

Given an integer probe, report whether that exact reading is filed anywhere in the archive.

Return true when the archive holds probe, otherwise false. The archive can be large, so your lookup must run in time logarithmic in the total reading count m * n; walking every page is too slow.

Examples

Example 1

Input: pages = [[-9, -4, 1], [6, 11, 20]], probe = 11
Output: true

Reading 11 sits in the second slot of the second page, so the archive holds it.

Example 2

Input: pages = [[-9, -4, 1], [6, 11, 20]], probe = 5
Output: false

Nothing between 1 and 6 was ever filed, so 5 is absent even though it falls inside the archive's range.

Example 3

Input: pages = [[-10], [0], [10]], probe = 10
Output: true

Pages of a single reading are allowed; the last page carries 10.

Constraints

m == pages.length
n == pages[i].length
1 <= m, n <= 100
-10^4 <= pages[i][j], probe <= 10^4
Readings ascend across the whole archive when the pages are read in order

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 locate_offset(pages: list[list[int]], probe: int) -> bool:

Java

public boolean locateOffset(int[][] pages, int probe)

The intended approach

Time O(log(m * n))Space O(1)

Because the pages ascend and no page overlaps the next, reading the archive page by page gives one strictly increasing sequence of length m * n. Binary search that virtual sequence and translate an index back with a division and a remainder by the page width, so no flattened copy is ever built.

Pythonentry point: locate_offset

def locate_offset(pages, probe):
    rows = len(pages)
    cols = len(pages[0])
    lo, hi = 0, rows * cols - 1
    while lo <= hi:
        mid = (lo + hi) // 2
        value = pages[mid // cols][mid % cols]
        if value == probe:
            return True
        if value < probe:
            lo = mid + 1
        else:
            hi = mid - 1
    return False

The obvious version

Scan every reading on every page and compare it against the probe, O(m * 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_locate_offset(pages, probe):
    for page in pages:
        for value in page:
            if value == probe:
                return True
    return False

How this problem was checked

Reference solution executed against 14 test cases — 14 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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