0117MediumArrayTwo PointersSweep Line

Shared Observation Windows

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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.

Two telescopes, one under the dome and one out on the ridge, each publish tonight's clear-sky windows as minute marks counted from dusk. dome and ridge hold those windows as [opens, closes] pairs, already ordered by opening mark, and no two windows in the same schedule ever touch: every window closes strictly before the next one in that schedule opens.

A paired reading needs both telescopes clear at once. Report every stretch of time that both schedules cover, again as [opens, closes] pairs ordered by opening mark. A stretch that comes down to a single minute mark, where opens equals closes, still counts as coverage and must be reported.

Either schedule may be empty, though not both. When the two schedules never coincide, return an empty list.

Examples

Example 1

Input: dome = [[2, 6], [10, 14]], ridge = [[6, 9], [13, 20]]
Output: [[6, 6], [13, 14]]

The first two windows meet only at mark 6, which still counts, and the later pair overlaps from mark 13 until the dome window closes at 14.

Example 2

Input: dome = [[2, 20]], ridge = [[3, 5], [8, 9], [12, 25]]
Output: [[3, 5], [8, 9], [12, 20]]

One long dome window swallows the first two ridge windows whole and clips the third at mark 20.

Example 3

Input: dome = [[1, 2]], ridge = [[3, 4]]
Output: []

The dome window has already closed before the ridge window opens, so nothing is covered twice.

Constraints

0 <= dome.length, ridge.length <= 1000
dome.length + ridge.length >= 1
0 <= opens < closes <= 10^9 for every window
Within one schedule the windows are sorted and disjoint: each window's closing mark is strictly less than the next window's opening mark.

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 shared_windows(dome: list[list[int]], ridge: list[list[int]]) -> list[list[int]]:

Java

public int[][] sharedWindows(int[][] dome, int[][] ridge)

The intended approach

Time O(m + n)Space O(1) beyond the reported list

Because each schedule is sorted and disjoint, one cursor per schedule is enough. The stretch two windows share runs from the later of their opening marks to the earlier of their closing marks, and it is real coverage whenever that span has not gone backwards. Then advance the cursor on whichever window closes first, since that window can share nothing with any later window in the other schedule.

Pythonentry point: shared_windows

def shared_windows(dome, ridge):
    shared = []
    i = 0
    j = 0
    while i < len(dome) and j < len(ridge):
        opens = dome[i][0] if dome[i][0] > ridge[j][0] else ridge[j][0]
        closes = dome[i][1] if dome[i][1] < ridge[j][1] else ridge[j][1]
        if opens <= closes:
            shared.append([opens, closes])
        if dome[i][1] < ridge[j][1]:
            i += 1
        else:
            j += 1
    return shared

The obvious version

Try every dome window against every ridge window, keep the overlaps, and sort what is left. 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_shared_windows(dome, ridge):
    found = []
    for a in dome:
        for b in ridge:
            opens = max(a[0], b[0])
            closes = min(a[1], b[1])
            if opens <= closes:
                found.append([opens, closes])
    found.sort()
    return found

How this problem was checked

Reference solution executed against 11 test cases — 11 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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