This is a Leetcode problem –

A city’s skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Now suppose you aregiven the locations and height of all the buildingsas shown on a cityscape photo (Figure A), write a program tooutput the skylineformed by these buildings collectively (Figure B).

The geometric information of each building is represented by a triplet of integers`[Li, Ri, Hi]`

, where`Li`

and`Ri`

are the`x`

coordinates of the left and right edge of the`i`

th building, respectively, and`Hi`

is its height. It is guaranteed that`0 ≤ Li, Ri ≤ INT_MAX`

,`0 < Hi ≤ INT_MAX`

, and`Ri - Li > 0`

. You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height`0`

.

For instance, the dimensions of all buildings in Figure A are recorded as:`[[2, 9, 10], [3, 7, 15], [5, 12, 12], [15, 20, 10], [19, 24, 8]]`

.

The output is a list of “key points” (red dots in Figure B) in the format of`[[x1,y1], [x2, y2], [x3, y3], ...]`

that uniquely defines a skyline.A key point is the left endpoint of a horizontal line segment. Note that the last key point, where the rightmost building ends, is merely used to mark the termination of the skyline, and always has zero height. Also, the ground in between any two adjacent buildings should be considered part of the skyline contour.

For instance, the skyline in Figure B should be represented as:`[[2, 10], [3, 15], [7, 12], [12, 0], [15, 10], [20, 8], [24, 0]]`

.

Notes –

The number of buildings in any input list is guaranteed to be in the range`[0, 10000]`

.

The input list is already sorted in ascending order by the left`x`

position`Li`

.

The output list must be sorted by the`x`

position.

There must be no consecutive horizontal lines of equal height in the output skyline. For instance,`[...[2 3], [4 5], [7 5], [11 5], [12 7]...]`

is not acceptable; the three lines of height 5 should be merged into one in the final output as such:`[...[2 3], [4 5], [12 7], ...]`

.

Here is my solution to this task using divide and conquer (in Python) –

`class Solution: def get_skyline(self, buildings): """ :type buildings: List[List[int]] :rtype: List[List[int]] """ if not buildings: return [] if len(buildings) == 1: return [[buildings[0][0], buildings[0][2]], [buildings[0][1], 0]] mid = len(buildings) // 2 left = self.get_skyline(buildings[:mid]) right = self.get_skyline(buildings[mid:]) return self.merge(left, right) def merge(self, left, right): h1, h2 = 0, 0 i, j = 0, 0 result = [] while i < len(left) and j < len(right): if left[i][0] < right[j][0]: h1 = left[i][1] corner = left[i][0] i += 1 elif right[j][0] < left[i][0]: h2 = right[j][1] corner = right[j][0] j += 1 else: h1 = left[i][1] h2 = right[j][1] corner = right[j][0] i += 1 j += 1 if self.is_valid(result, max(h1, h2)): result.append([corner, max(h1, h2)]) result.extend(right[j:]) result.extend(left[i:]) return result def is_valid(self, result, new_height): return not result or result[-1][1] != new_height `

Here is a sample input/output –

`output = Solution() print(output.get_skyline([[2, 9, 10], [3, 7, 15], [5, 12, 12], [15, 20, 10], [19, 24, 8]])) >>> [[2, 10], [3, 15], [7, 12], [12, 0], [15, 10], [20, 8], [24, 0]] `

Here is the time taken for this output –

`%timeit output.get_skyline([[2, 9, 10], [3, 7, 15], [5, 12, 12], [15, 20, 10], [19, 24, 8]]) >>> 27.6 µs ± 3.65 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each) `

So, I would like to know whether I could make this program more efficient and/or shorter. Any alternatives are welcome.