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Home Code Solutions

The Skyline Problem – LeetCode Solution

The Skyline Problem - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 5, 2022
Reading Time: 3 mins read
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Leetcode All Problems Solutions

Leetcode All Problems Solutions

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Table of Contents

  • The Skyline Problem – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ The Skyline Problem LeetCode Solution
  • Java The Skyline Problem LeetCode Solution
  • Python 3 The Skyline Problem LeetCode Solution
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The Skyline Problem – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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]]
Explanation:
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]]

 

Constraints:

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

C++ The Skyline Problem LeetCode Solution


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class Solution {
public:
    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]:liveBlg.top().second; // 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() && ( liveBlg.top().second <= cur_X) ) liveBlg.pop();
            }
            else
            { // 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++;
                }
            }
            cur_H = liveBlg.empty()?0:liveBlg.top().first; // outut the top one
            if(res.empty() || (res.back().second != cur_H) ) res.push_back(make_pair(cur_X, cur_H));
        }
        return res;
    }
};

Java The Skyline Problem LeetCode Solution


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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));
    pq.offer(0);
    int prev = 0;
    for(int[] h:height) {
        if(h[1] < 0) {
            pq.offer(-h[1]);
        } else {
            pq.remove(h[1]);
        }
        int cur = pq.peek();
        if(prev != cur) {
            result.add(new int[]{h[0], cur});
            prev = cur;
        }
    }
    return result;
}

 



Python 3 The Skyline Problem LeetCode Solution


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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
                heapq.heappop(max_heap)
            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:] 



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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Tags: Cc++14full solutionGojavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionpypy 3Python 2python 3rubyrustSolutionThe Skyline Problem
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