'''
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], ...]
'''

from heapq import *

class Solution:
    # @param {integer[][]} buildings
    # @return {integer[][]}
    def getSkyline(self, buildings):
        if not buildings: return [] 
        skyline = []
        i = 0; n = len(buildings)
        liveHR = []
        
        while i < n or liveHR:
            if not liveHR or (i < n and buildings[i][0] <= -liveHR[0][1]):
                x = buildings[i][0]
                while i < n and buildings[i][0] == x:
                    heappush(liveHR, (-buildings[i][2], -buildings[i][1]))
                    i += 1
            else:
                x = -liveHR[0][1]
                while liveHR and -liveHR[0][1] <= x:
                    heappop(liveHR)
            height = len(liveHR) and -liveHR[0][0]
            if not skyline or height != skyline[-1][1]:
                skyline += [x, height],
        
        return skyline 
        
# https://leetcode.com/discuss/37736/108-ms-17-lines-body-explained
# https://leetcode.com/discuss/37630/my-c-code-using-one-priority-queue-812-ms


@@ -1970,16 +1970,945 @@
 .%5D%0A'''%0A%0A
+'''%0Aanalysis:%0AThe idea is to do line sweep and just process the buildings only at the start and end points. %0AThe key is to use a priority queue to save all the buildings that are still %22alive%22. %0AThe queue is sorted by its height and end time (the larger height first and if equal height, %0Athe one with a bigger end time first). For each iteration, we first find the current process time, %0Awhich is either the next new building start time or the end time of the top entry of the live queue. %0AIf the new building start time is larger than the top one end time, then process the one in the queue %0Afirst (pop them until it is empty or find the first one that ends after the new building); otherswise, %0Aif the new building starts before the top one ends, then process the new building (just put them in the queue). %0AAfter processing, output it to the resulting vector if the height changes. Complexity is the worst case O(NlogN)%0A'''%0A%0A
 from hea

Update The_Skyline_Problem.py