Q218. The Skyline 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 are given the locations and height of all the buildings as shown on a cityscape photo (Figure A), write a program to output the skyline formed by these buildings collectively (Figure B).

Buildings Skyline Contour
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 ith 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], ...]

I struggled quite a bit with this problem. I was not quite sure if I should use heap and I got stuck on how to update the current height when I hit the end of a building (especially when the current height is given by that building).

It turns out that we could use heap to maintain the current height. The key observation is that the current height is the maximum amount all the buildings whose end position (right anchor) is to the right of the current position. So, we could build a heap to maintain the current height by filtering out maximum heights that passed the current position.

OK, once we have the current height, the algorithm is more straightforward to describe.

• At each starting point of the building, determine if its height is higher than the current height. If so, it will contribute to the skyline.
• At each end point of the building, determine if its height is higher than the current height (after removing this building). If so, this building will contribute to the skyline.

There is also corner case about how to handle buildings side-by-side with the same height. And the situation where building are at exactly the same start and end positions but different height. And we need to consider higher building before the lower. And we need to consider the start position before the end position (if they are at the same position).
And when we add a new point, we need to make sure that it has different height than the one before it.

The Python submission.

class Solution:
    def getSkyline(self, buildings: List[List[int]]) -> List[List[int]]:
        points = []
        for i, j, h in buildings:
            points.append([i, h, -1, j])
            points.append([j, h, 1, -1])
        points = sorted(points, key=lambda x: (x[0], x[2], -x[1]))
        height_mem = []
        curr_height = 0
        res = []
        for pos, height, type_, end in points:
            while len(height_mem) > 0 and height_mem[0][1] <= pos:
                heapq.heappop(height_mem)
            if len(height_mem) > 0:
                curr_height = -height_mem[0][0]
            else:
                curr_height = 0
            if type_ == -1:
                if height > curr_height and (len(res) == 0 or res[-1][1] != height):
                    res.append((pos, height))
                heapq.heappush(height_mem, (-height, end))
            elif type_ == 1:
                if curr_height < height and (len(res) == 0 or res[-1][1] != curr_height):
                    res.append((pos, curr_height))
        return res