0218. The Skyline Problem

https://leetcode.com/problems/the-skyline-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 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.

ac

class Solution {
    public List<List<Integer>> getSkyline(int[][] buildings) {
        List<List<Integer>> res  = new ArrayList<>();
        // edge cases
        if (buildings == null || buildings.length == 0) return res;
        
        List<Point> points = new ArrayList<>();
        for (int[] b : buildings) {
            points.add(new Point(b[0], b[2], true)); // start point
            points.add(new Point(b[1], b[2], false)); // end point
        }
        Collections.sort(points);
        
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a); // record heights
        pq.offer(0);  // When there is no building, height is 0.
        int highest = pq.peek();
        
        for (Point p : points) {
            if (p.isStart) {
                // Meet start point, we got a new building, add its height.
                pq.offer(p.height);
            } else {
                // Meet end point, we pass this building, remove its height.
                pq.remove(p.height);
            }
            
            int newHighest = pq.peek();
            if (newHighest != highest) {  
                // The height is changed after we add or remove a building. Record this point.
                res.add(Arrays.asList(p.index, newHighest));
                highest = newHighest;
            }
        }
        
        return res;
    }
    
    class Point implements Comparable<Point> {
        int index;
        int height;
        boolean isStart;
        
        Point(int index, int height, boolean isStart) {
            this.index = index;
            this.height = height;
            this.isStart = isStart;
        }
        
        @Override
        public int compareTo(Point other) {
            if (this.index == other.index) {
                int thisHeight = this.height;
                // This is tricky: if these are same start points, handle heighest first.
                // If they are end points, handle lowest first. 
                // Becase if we remove heighest, the heigh changed and we record a point we don't want.
                // example: [[4,9,10],[4,9,15],[4,9,12]]
                // So, for all points in the same index, the process order is: high->low start points, then low->high end points
                if (this.isStart) {
                    if (other.isStart) {
                        return other.height - this.height; // Descending height for start points
                    } else {
                        return -1; // always handle start point first
                    }
                } else {
                    if (other.isStart) {
                        return 1; // always handle start point first
                    } else {
                        return this.height - other.height; // Ascending height for end points
                    }
                }
            } 
            
            return this.index - other.index;
        }
    }
}

// O(NlogN) time, O(N) space
// The key is to record the point when height is changed after we add or remove a building. The tricky part is the ordering for all points in the same index.