# [leetcode] The Skyline Problem

### 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).

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], ...]`

Credits:
Special thanks to @stellari for adding this problem, creating these two awesome images and all test cases.

https://briangordon.github.io/2014/08/the-skyline-problem.html

Store the points (left-top, right-top) and sort it according to their x coordinate.

Store left point and point separately. Here I convert the height of right point to negative to differentiate them.

What’s more, in order to avoid ambiguous when height is 0, height of right point is subtracted by 1.

So that all heights >= 0 belongs to left point, all heights < 0 belongs to right point.

Imagine a vertical line scans the whole buildings from left to right.

maxHeight stores the current alive building, which means the height of final graph is the same as the alive building.

When a left point comes, push it in to heap, add its count in hash map by 1.

When a right point comes, minus its count in hash map by 1,

```class Solution {
public:
vector<pair<int, int>> getSkyline(vector<vector<int>>& buildings) {
priority_queue<int> maxHeight;
unordered_map<int, int> map;//count heights
vector<pair<int, int>> points;
vector<pair<int, int>> ans;
for(auto building: buildings){
points.push_back(make_pair(building, building));//left point
points.push_back(make_pair(building, -building - 1));//right point, in case height is 0, distinguish it
}
sort(points.begin(), points.end(), [](pair<int, int>& a, pair<int, int>& b){
return a.first < b.first;
});
maxHeight.push(0);
map = 1;
int prevHeight = -1;
for(int i = 0; i < points.size(); i++){
int x = points[i].first;
int height = points[i].second;
if(height >= 0){
//left point, push it into queue
maxHeight.push(height);
if(map.find(height) == map.end()) map[height] = 1;
else map[height]++;
}
else {
//right point, erase it from queue
height = -height - 1;
map[height]--;
}
if(i < points.size() - 1 && points[i + 1].first == x){
continue;//not the last point and the x coordinate of next point is same as current point
}
//remove erased elements
while(map[maxHeight.top()] == 0){
maxHeight.pop();
}

if(prevHeight == maxHeight.top()) continue;

ans.push_back(make_pair(x, maxHeight.top()));
prevHeight = maxHeight.top();
}
return ans;

}
};```

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