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Leetcode The Skyline problem solution

YASH PAL, 31 July 202420 January 2026

In this Leetcode The Skyline problem solution 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.

Leetcode The Skyline problem solution

Leetcode The Skyline problem solution in Python.

class Solution:
    def getSkyline(self, buildings: List[List[int]]) -> List[List[int]]:
        if not buildings:
            return []
        points = []
        for left, right, height in buildings:
            points.append((left, -height))
            points.append((right, height))
        points = sorted(points, key=lambda x: (x[0], x[1])) # break ties when x coordiantes are equal
        from heapq import heappop, heappush, heapify
        res, pq = [], [float('inf')]
        prev_max_height = 0
        for x, height in points:
            if height < 0:
                heappush(pq, height)
            else:
                pq.remove(-height)
                heapify(pq)
            cur_max_height = pq[0]
            if cur_max_height != prev_max_height:
                if cur_max_height == float('inf'):
                    cur_max_height = 0
                res.append([x, -cur_max_height])
                prev_max_height = cur_max_height
        return res

The Skyline problem solution in Java.

class cord{
    int x,y;
    boolean isStart;
    cord(int x,int y,boolean f){
        this.x=x;this.y=y;isStart=f;
    }
}
class Solution {
    public List<List<Integer>> getSkyline(int[][] buildings) {
        List<List<Integer>> ans=new ArrayList<>();
        if(buildings.length==0) return ans;
        List<cord> l=new ArrayList<>();
        for(int i[]:buildings){
            cord c=new cord(i[0],i[2],true),c1=new cord(i[1],i[2],false);
            l.add(c);l.add(c1);
        }
        Collections.sort(l,new Comparator<cord>() {
            @Override
            public int compare(cord a,cord b){
                if(a.x!=b.x)    return a.x-b.x;
                else{
                    if(a.isStart && b.isStart) return b.y-a.y;
                    else if(a.isStart!=b.isStart)  return a.isStart?-1:1;
                    else    return a.y-b.y;
                }
            }
        });
        TreeMap<Integer,Integer> m=new TreeMap<>();
        m.put(0,1);
        int prevMax=0;
        for(cord i:l){
            if(i.isStart)  m.put(i.y,m.getOrDefault(i.y,0)+1);
            else{
                int v=m.get(i.y)-1;
                if(v==0)   m.remove(i.y);
                else   m.put(i.y,v);
            }
            int max=m.lastKey();
            if(max!=prevMax){
                List<Integer> t=new ArrayList<>();
                t.add(i.x);t.add(max);
                ans.add(t);
                prevMax=max;
            }
        }
        return ans;
    }
}

Problem solution in C++.

bool cmp(vector<int> &A, vector<int> &B){
    if(A[0]!=B[0])
        return A[0]<B[0];
        else if(A[1]==B[1])
            return B[2]==1; 
    else if(A[2]==0 && B[2]==0)
        return A[1]>B[1];
    else if(A[2]==1 && B[2]==1)
        return A[1]<B[1];
    return true;
}
class Solution {
public:
    vector<vector<int>> getSkyline(vector<vector<int>>& buildings) {
        int n=buildings.size();
        
        vector< vector<int> > A;
        for(int i=0;i<buildings.size();i++){
            A.push_back({buildings[i][0],buildings[i][2],0});
            A.push_back({buildings[i][1],buildings[i][2],1});
        }
        sort(A.begin(),A.end(),cmp);
        multiset<int> pq;
        vector< vector<int> > ans;
        int maxValue=0;
        pq.insert(0);
        for(int i=0;i<A.size();i++){
            if(A[i][2]==0){
                pq.insert(A[i][1]);
            }
            else{
                pq.erase(pq.find(A[i][1]));
            }
            auto itr=pq.end();
            itr--;
            if(maxValue!=*itr)
            ans.push_back({A[i][0],*itr});
            maxValue=*itr;
        }
        return ans;
    }
};

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