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The Sweep Line algorithm involves moving a vertical line from left to right over the x-coordinates and maintaining a list of currently active buildings using a max-heap. The priority queue helps keep track of the tallest building at each point. As the line reaches the start of a new building, add it to the heap, and as it reaches the end of a building, remove it. The key points are detected based on changes in the maximum height at points where buildings start or end.
Time Complexity: O(N log N), where N is the number of events (two for each building).
Space Complexity: O(N), as we store up to N building heights in the heap.
1#include <vector>
2#include <set>
3#include <algorithm>
4using namespace std;
5
6vector<vector<int>> getSkyline(vector<vector<int>>& buildings) {
7 vector<pair<int, int>> events;
8 for (auto& b : buildings) {
9 events.emplace_back(b[0], -b[2]);
10 events.emplace_back(b[1], b[2]);
11 }
12 sort(events.begin(), events.end());
13
14 multiset<int> heights = {0};
15 vector<vector<int>> res;
16 int prev = 0;
17
18 for (const auto& e : events) {
19 if (e.second < 0) {
20 heights.insert(-e.second);
21 } else {
22 heights.erase(heights.find(e.second));
23 }
24 int current = *heights.rbegin();
25 if (current != prev) {
26 res.push_back({e.first, current});
27 prev = current;
28 }
29 }
30
31 return res;
32}This C++ solution employs a multiset to track the heights of buildings, similar to using a priority queue. By managing events and ensuring at each point the skyline height reflects the maximum possible height, the code outputs the required key points only when there's a change.
This approach involves breaking down the problem using the divide and conquer strategy. The buildings array is split into two halves recursively. The base case is a single building, where the skyline can directly be calculated. The results are then merged, taking care of overlapping heights and ensuring continuity and distinct key points.
Time Complexity: O(N log N), due to division of buildings and merging.
Space Complexity: O(N), needing space for recursive stack and individual skyline lists.
1def
This Python code uses the divide and conquer strategy where each call splits the building list into two halves and recursively solves for both sides, merging their results into a coherent skyline outline by comparing the heights at each step and ensuring the output is minimal.