The Skyline Problem
Problem Description​
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 buildings are represented by a list of triplets [left, right, height], where:
leftis the x coordinate of the left edge of the building.rightis the x coordinate of the right edge of the building.heightis the height of the building.
The output is a list of "key points" (coordinates) that uniquely define 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 not included.
Example​
Example 1:
Input: buildings = [[2,9,10],[3,7,15],[5,12,12],[15,20,10],[19,24,8]]
Output: [[2,10],[3,15],[7,12],[12,0],[15,10],[20,8],[24,0]]
Example 2:
Input: buildings = [[0,2,3],[2,5,3]]
Output: [[0,3],[5,0]]
Constraints​
1 <= buildings.length <= 10^40 <= left_i < right_i <= 2^31 - 11 <= height_i <= 2^31 - 1buildingsis sorted byleftin non-decreasing order.
Approach​
Approach Overview​
To solve the skyline problem, we can use a combination of a line sweep algorithm and a max-heap to track the current heights of buildings as we move from left to right across the x-axis.
Detailed Steps​
-
Extract Key Events:
- For each building, extract two events: one for the start (
left) of the building with its height as positive, and one for the end (right) of the building with its height as negative.
- For each building, extract two events: one for the start (
-
Sort Events:
- Sort the events primarily by the x-coordinate. If two events have the same x-coordinate, process the event with the taller height first.
-
Process Events Using a Max-Heap:
- Use a max-heap to keep track of the current building heights as we process each event.
- For each event:
- If it's a start event (height is positive), add the height to the heap.
- If it's an end event (height is negative), remove the height from the heap.
- After processing the event, check the current highest height in the heap. If it has changed from the previous highest height, add a new key point to the result.
Code Examples​
Python​
from heapq import heappush, heappop
class Solution:
def getSkyline(self, buildings: List[List[int]]) -> List[List[int]]:
events = []
for left, right, height in buildings:
events.append((left, -height, right))
events.append((right, 0, 0))
events.sort()
result, heap = [[0, 0]], [(0, float('inf'))]
for x, negH, R in events:
while x >= heap[0][1]:
heappop(heap)
if negH:
heappush(heap, (negH, R))
if result[-1][1] != -heap[0][0]:
result.append([x, -heap[0][0]])
return result[1:]
C++​
class Solution {
public:
vector<vector<int>> getSkyline(vector<vector<int>>& buildings) {
vector<pair<int, int>> events;
for (const auto& building : buildings) {
events.emplace_back(building[0], -building[2]);
events.emplace_back(building[1], building[2]);
}
sort(events.begin(), events.end());
vector<vector<int>> result;
multiset<int> heights = {0};
int prevMaxHeight = 0;
for (const auto& event : events) {
int x = event.first;
int h = event.second;
if (h < 0) {
heights.insert(-h);
} else {
heights.erase(heights.find(h));
}
int currentMaxHeight = *heights.rbegin();
if (currentMaxHeight != prevMaxHeight) {
result.push_back({x, currentMaxHeight});
prevMaxHeight = currentMaxHeight;
}
}
return result;
}
};
Java​
class Solution {
public List<List<Integer>> getSkyline(int[][] buildings) {
List<int[]> events = new ArrayList<>();
for (int[] building : buildings) {
events.add(new int[]{building[0], -building[2]});
events.add(new int[]{building[1], building[2]});
}
Collections.sort(events, (a, b) -> {
if (a[0] != b[0]) return Integer.compare(a[0], b[0]);
return Integer.compare(a[1], b[1]);
});
List<List<Integer>> result = new ArrayList<>();
PriorityQueue<Integer> heights = new PriorityQueue<>(Collections.reverseOrder());
heights.add(0);
int prevMaxHeight = 0;
for (int[] event : events) {
if (event[1] < 0) {
heights.add(-event[1]);
} else {
heights.remove(event[1]);
}
int currentMaxHeight = heights.peek();
if (currentMaxHeight != prevMaxHeight) {
result.add(Arrays.asList(event[0], currentMaxHeight));
prevMaxHeight = currentMaxHeight;
}
}
return result;
}
}
C​
#include <stdlib.h>
typedef struct {
int x, h;
} Event;
int cmp(const void *a, const void *b) {
Event *ea = (Event *)a;
Event *eb = (Event *)b;
if (ea->x != eb->x) return ea->x - eb->x;
return ea->h - eb->h;
}
int* getSkyline(int** buildings, int buildingsSize, int* buildingsColSize, int* returnSize) {
int eventCount = 2 * buildingsSize;
Event *events = (Event *)malloc(eventCount * sizeof(Event));
int *heights = (int *)malloc(eventCount * sizeof(int));
int *result = (int *)malloc(eventCount * 2 * sizeof(int));
int heightsSize = 0, resultSize = 0, maxHeight = 0;
for (int i = 0; i < buildingsSize; ++i) {
events[2 * i].x = buildings[i][0];
events[2 * i].h = -buildings[i][2];
events[2 * i + 1].x = buildings[i][1];
events[2 * i + 1].h = buildings[i][2];
}
qsort(events, eventCount, sizeof(Event), cmp);
heights[heightsSize++] = 0;
for (int i = 0; i < eventCount; ++i) {
Event event = events[i];
if (event.h < 0) {
heights[heightsSize++] = -event.h;
} else {
for (int j = 0; j < heightsSize; ++j) {
if (heights[j] == event.h) {
for (int k = j; k < heightsSize - 1; ++k) {
heights[k] = heights[k + 1];
}
--heightsSize;
break;
}
}
}
int currentMaxHeight = 0;
for (int j = 0; j < heightsSize; ++j) {
if (heights[j] > currentMaxHeight) currentMaxHeight = heights[j];
}
if (currentMaxHeight != maxHeight) {
result[resultSize++] = event.x;
result[resultSize++] = currentMaxHeight;
maxHeight = currentMaxHeight;
}
}
*returnSize = resultSize;
free(events);
free(heights);
return result;
}
Complexity​
- Time Complexity:
O(n log n), wherenis the number of buildings. Sorting the events takesO(n log n)time, and processing each event takesO(log n)time due to heap operations. - Space Complexity:
O(n), due to the storage of events and the heap structure.