The Skyline Problem LeetCode Solution

Problem – The Skyline Problem

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],...]

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]]
Figure A shows the buildings of the input.
Figure B shows the skyline formed by those buildings. The red points in figure B represent the key points in the output list.

Example 2:

Input: buildings = [[0,2,3],[2,5,3]]
Output: [[0,3],[5,0]]


  • 1 <= buildings.length <= 104
  • 0 <= lefti < righti <= 231 - 1
  • 1 <= heighti <= 231 - 1
  • buildings is sorted by lefti in non-decreasing order.

The Skyline Problem LeetCode Solution in Java

	public List<int[]> getSkyline(int[][] buildings) {
    List<int[]> result = new ArrayList<>();
    List<int[]> height = new ArrayList<>();
    for(int[] b:buildings) {
        height.add(new int[]{b[0], -b[2]});
        height.add(new int[]{b[1], b[2]});
    Collections.sort(height, (a, b) -> {
            if(a[0] != b[0]) 
                return a[0] - b[0];
            return a[1] - b[1];
    Queue<Integer> pq = new PriorityQueue<>((a, b) -> (b - a));
    int prev = 0;
    for(int[] h:height) {
        if(h[1] < 0) {
        } else {
        int cur = pq.peek();
        if(prev != cur) {
            result.add(new int[]{h[0], cur});
            prev = cur;
    return result;

The Skyline Problem LeetCode Solution in Python

class Solution:
    def getSkyline(self, buildings: List[List[int]]) -> List[List[int]]:
        # for the same x, (x, -H) should be in front of (x, 0)
        # For Example 2, we should process (2, -3) then (2, 0), as there's no height change
        x_height_right_tuples = sorted([(L, -H, R) for L, R, H in buildings] + [(R, 0, "doesn't matter") for _, R, _ in buildings])   
        # (0, float('inf')) is always in max_heap, so max_heap[0] is always valid
        result, max_heap = [[0, 0]], [(0, float('inf'))]
        for x, negative_height, R in x_height_right_tuples:
            while x >= max_heap[0][1]:
                # reduce max height up to date, i.e. only consider max height in the right side of line x
            if negative_height:
                # Consider each height, as it may be the potential max height
                heapq.heappush(max_heap, (negative_height, R))
            curr_max_height = -max_heap[0][0]
            if result[-1][1] != curr_max_height:
                result.append([x, curr_max_height])
        return result[1:] 

The Skyline Problem LeetCode Solution in C++

class Solution {
    vector<pair<int, int>> getSkyline(vector<vector<int>>& buildings) {
        vector<pair<int, int>> res;
        int cur=0, cur_X, cur_H =-1,  len = buildings.size();
        priority_queue< pair<int, int>> liveBlg; // first: height, second, end time
        while(cur<len || !liveBlg.empty())
        { // if either some new building is not processed or live building queue is not empty
            cur_X = liveBlg.empty()? buildings[cur][0]; // next timing point to process

            if(cur>=len || buildings[cur][0] > cur_X)
            { //first check if the current tallest building will end before the next timing point
                  // pop up the processed buildings, i.e. those  have height no larger than cur_H and end before the top one
                while(!liveBlg.empty() && ( <= cur_X) ) liveBlg.pop();
            { // if the next new building starts before the top one ends, process the new building in the vector
                cur_X = buildings[cur][0];
                while(cur<len && buildings[cur][0]== cur_X)  // go through all the new buildings that starts at the same point
                {  // just push them in the queue
                    liveBlg.push(make_pair(buildings[cur][2], buildings[cur][1]));
            cur_H = liveBlg.empty()?; // outut the top one
            if(res.empty() || (res.back().second != cur_H) ) res.push_back(make_pair(cur_X, cur_H));
        return res;
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