218. The Skyline Problem¶

Problem Description¶

LeetCode Problem 218: A city's skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Given the locations and heights of all the buildings, return the skyline formed by these buildings collectively.

The geometric information of each building is given in the array buildings where buildings[i] = [lefti, righti, heighti]:

lefti is the x coordinate of the left edge of the ith building.
righti is the x coordinate of the right edge of the ith building.
heighti is the height of the ith building.

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

The skyline should be represented as a list of "key points" sorted by their x-coordinate in the form [[x1,y1],[x2,y2],...]. Each key point is the left endpoint of some horizontal segment in the skyline except the last point in the list, which always has a y-coordinate 0 and is used to mark the skyline's termination where the rightmost building ends. Any ground between the leftmost and rightmost buildings should be part of the skyline's contour.

Note: 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],...]

Clarification¶

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Assumption¶

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Solution¶

Approach 1: Sweep Line with Max-Heap¶

The problem can be solved using sweep line algorithm combined with a max-heap. It processes building edges in order and tracks active building heights to determine the skyline.

Python

import heapq

class Solution:
    def getSkyline(self, buildings: list[list[int]]) -> list[list[int]]:
        # Step 1: Create all the critical points
        events = []
        for left, right, height in buildings:
            events.append((left, -height))  # entering event (using negative height)
            events.append((right, height))  # exiting event

        # Step 2: Sort events
        # Sort by x-coordinate; if tie, sort by height: entering before exiting
        events.sort(key=lambda x: (x[0], x[1]))

        # Step 3: Max-heap to keep track of active building heights
        result = []
        max_heap = [0]  # initial ground height
        heap_counter = {0: 1}  # track occurrences
        prev_max = 0

        for x, h in events:
            if h < 0:  # entering a building
                heapq.heappush(max_heap, h)
                heap_counter[-h] = heap_counter.get(-h, 0) + 1
            else:  # exiting a building
                heap_counter[h] -= 1

            # Clean up the heap (remove heights that are no longer active)
            while max_heap and heap_counter[-max_heap[0]] == 0:
                heapq.heappop(max_heap)

            current_max = -max_heap[0]
            if current_max != prev_max:
                result.append([x, current_max])
                prev_max = current_max

        return result

Complexity Analysis of Approach 1¶

Time complexity: \(O(n \log n)\)
- Sorting takes \(O(n \log n)\) time for \(n\) events (twice the number of buildings).
- Go through \(n\) events and each iteration takes \(O(\log n)\) heap operation. So the time complexity is \(O(n \log n)\)
- So the overall time complexity is \(O(n \log n)\)
Space complexity: \(O(n)\)
- Sorting events takes \(O(n)\) space.
- The heap takes \(O(n)\) space.
- So the over space complexity is \(O(n)\).

Approach 2:¶

Solution

python

code

Complexity Analysis of Approach 2¶

Time complexity: \(O(1)\)
Explanation
Space complexity: \(O(n)\)
Explanation

Comparison of Different Approaches¶

The table below summarize the time complexity and space complexity of different approaches:

Approach	Time Complexity	Space Complexity
Approach -	\(O(1)\)	\(O(n)\)
Approach -	\(O(1)\)	\(O(n)\)