Problem Statement
This task involves constructing a "skyline" from a series of buildings described by their geometric dimensions. Each building is depicted as a rectangle grounded on a flat surface and defined by three integers that represent its left and right x-coordinates and its height.
The goal is to determine the overall outline or skyline that this group of buildings forms when viewed from a distance. The output should be a list of "key points" that describe major changes in the height of this skyline, starting from the leftmost part of the first building to the rightmost edge of the last building, ensuring it always ends with the ground level. These key points are the coordinates where the height of the skyline changes. It is crucial to ensure that there are no redundant points where the skyline remains at the same height, to portray the outline as effectively as possible.
Examples
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 <= 1040 <= lefti < righti <= 231 - 11 <= heighti <= 231 - 1buildingsis sorted byleftiin non-decreasing order.
Approach and Intuition
Determine crucial
x-coordinates:- Start by identifying all the critical points which are all the left and right edges of the buildings. These points will help us script the potential positions where the skyline can change.
Use height priority:
- A priority queue can be employed to maintain the current highest height at surveyed points as we sweep from left to right across the skyline. The height changes either at the start or the end of a building.
Sweep horizontally:
- Initiate a vertical sweep-line technique. Move from left to right across the buildings' critical points, adjusting the skyline height as buildings start or stop contributing to the skyline.
Record changes:
- Each time the height changes from the previous height, record a new key point. This ensures we only capture the points at which the actual silhouette alters.
Efficient merging:
- Ensure efficient merger of consecutive segments at the same height by checking the current skyline height before recording a new key point.
By implementing this approach, we capture the skyline efficiently and portray a city's architectural contour accurately without redundancy. The solution leverages data structures like priority queues to handle the dynamically changing maximum building height as we progress through the x-coordinates. This approach is both intuitive and efficient in dealing with the complexities of varying building heights and widths.
Solutions
- C++
using IntMatrix = std::vector<std::vector<int>>;
class Skylines {
public:
IntMatrix calculateSkyline(IntMatrix& buildings) {
// Start recursive divide-and-conquer process on buildings
return splitAndMerge(buildings, 0, buildings.size() - 1);
}
IntMatrix splitAndMerge(IntMatrix& buildings, int start, int end) {
if (start == end) {
return {{buildings[start][0], buildings[start][2]}, {buildings[start][1], 0}};
}
int mid = start + (end - start) / 2;
IntMatrix skyLeft = splitAndMerge(buildings, start, mid);
IntMatrix skyRight = splitAndMerge(buildings, mid + 1, end);
return blendSkylines(skyLeft, skyRight);
}
IntMatrix blendSkylines(IntMatrix& left, IntMatrix& right) {
int lIndex = 0, rIndex = 0;
int heightLeft = 0, heightRight = 0;
int currentHeight = 0, xPosition = 0;
IntMatrix mergedSkyline;
while (lIndex < left.size() && rIndex < right.size()) {
int xLeft = left[lIndex][0], xRight = right[rIndex][0];
if (xLeft < xRight) {
xPosition = xLeft;
currentHeight = max(left[lIndex][1], heightRight);
heightLeft = left[lIndex++][1];
} else if (xLeft > xRight) {
xPosition = xRight;
currentHeight = max(right[rIndex][1], heightLeft);
heightRight = right[rIndex++][1];
} else {
xPosition = xLeft;
currentHeight = max(left[lIndex][1], right[rIndex][1]);
heightLeft = left[lIndex++][1];
heightRight = right[rIndex++][1];
}
if (mergedSkyline.empty() || currentHeight != mergedSkyline.back()[1]) {
mergedSkyline.push_back({xPosition, currentHeight});
}
}
while (lIndex < left.size()) {
mergedSkyline.push_back(left[lIndex++]);
}
while (rIndex < right.size()) {
mergedSkyline.push_back(right[rIndex++]);
}
return mergedSkyline;
}
};
The Skyline Problem is efficiently solved using a C++ program based on a divide and conquer approach combined with a merging strategy for combining skylines.
Essentially, the skyline problem involves computing the silhouette of a city defined by a number of rectangular buildings. The supplied C++ solution follows these principles:
- Recursive Splitting: Buildings are recursively divided into smaller subsets until individual buildings are isolated.
- Merging Skylines: Pairs of building subsets are merged back together. During the merge, comparisons between the heights of two skylines are made to construct a single, seamless skyline.
- Blend Skylines Function: This is the core functionality where the two skylines (left and right) are compared, and the points where the skyline needs to change in height are noted. The algorithm ensures that at any transition point, the height selected is the maximum visible height, which accurately forms the correct outline of the combined skylines.
The class Skylines encapsulates all the functionality, with its primary method, calculateSkyline, initiating the process using the auxiliary methods splitAndMerge and blendSkylines. The splitAndMerge function handles dividing the building list and invoking the merging of the divided parts.
By using vectors to store the skylines points and careful checks to add only necessary points (where the height of the skyline actually changes), the program ensures efficiency and clarity of the final skyline contour. Each function communicates through return types IntMatrix, which represents a list of skyline-defining points as pairs of X coordinates and heights. This strategic splitting and blending ensure that the solution is both scalable and manageable.
Consider leveraging this setup for solving similar problems involving merging of intervals or other geometrical figure blending challenges in graphical applications, simulation systems, or urban planning tools.
- Java
class Solution {
public List<List<Integer>> calculateSkyline(int[][] buildings) {
return divide(buildings, 0, buildings.length - 1);
}
public List<List<Integer>> divide(int[][] buildings, int start, int end) {
if (start == end) {
List<List<Integer>> skyPoints = new ArrayList<>();
skyPoints.add(Arrays.asList(buildings[start][0], buildings[start][2]));
skyPoints.add(Arrays.asList(buildings[start][1], 0));
return skyPoints;
}
int mid = (end - start)/2 + start;
List<List<Integer>> skylineLeft = divide(buildings, start, mid);
List<List<Integer>> skylineRight = divide(buildings, mid + 1, end);
return merge(skylineLeft, skylineRight);
}
public List<List<Integer>> merge(List<List<Integer>> skyline1, List<List<Integer>> skyline2) {
List<List<Integer>> finalSkyline = new ArrayList<>();
int i = 0, j = 0;
int height1 = 0, height2 = 0;
int currentX, currentHeight;
while (i < skyline1.size() && j < skyline2.size()) {
int point1X = skyline1.get(i).get(0);
int point2X = skyline2.get(j).get(0);
if (point1X < point2X) {
height1 = skyline1.get(i).get(1);
currentX = point1X;
currentHeight = Math.max(height1, height2);
i++;
} else if (point1X > point2X) {
height2 = skyline2.get(j).get(1);
currentX = point2X;
currentHeight = Math.max(height1, height2);
j++;
} else {
height1 = skyline1.get(i).get(1);
height2 = skyline2.get(j).get(1);
currentX = point1X;
currentHeight = Math.max(height1, height2);
i++;
j++;
}
if (finalSkyline.isEmpty() || finalSkyline.get(finalSkyline.size() - 1).get(1) != currentHeight) {
finalSkyline.add(Arrays.asList(currentX, currentHeight));
}
}
while (i < skyline1.size()) {
finalSkyline.add(skyline1.get(i));
i++;
}
while (j < skyline2.size()) {
finalSkyline.add(skyline2.get(j));
j++;
}
return finalSkyline;
}
}
The "Skyline Problem" solution calculates a skyline view from a series of rectangle buildings represented by their bottom-left and top-right coordinates. The Java implementation employs a divide-and-conquer approach.
- The
calculateSkylinemethod begins by initiating the divide strategy over the entire range of building coordinates. - The
dividefunction recursively splits the building list until individual buildings remain, creating skylines for one building at a time. Then, recursively merges these skylines using themergefunction. - For a single building, the
dividemethod provides the x-coordinate of its left edge, its y-coordinate (height), and the right edge where the height drops back to zero. - The
mergefunction compares the edges of two skyline lists (left and right). It updates the current maximum height by picking the highest point visible at any x-coordinate and resolves conflicts where both skylines share the same x-coordinate. - It maintains a list of the resulting merged skylines ensuring no two consecutive points share the same height to avoid unnecessary points in the final skyline curve.
mergeiteratively adds remaining points from either skyline should one list be exhausted before the other.
This method efficiently assembles an accurate skyline by breaking the problem into simpler sub-problems and merging them optimally.
- Python
class Solution:
def computeSkyline(self, buildings: List[List[int]]) -> List[List[int]]:
count = len(buildings)
if count == 0:
return []
if count == 1:
return [[buildings[0][0], buildings[0][2]], [buildings[0][1], 0]]
left_part = self.computeSkyline(buildings[: count // 2])
right_part = self.computeSkyline(buildings[count // 2:])
return self.combineSkylines(left_part, right_part)
def combineSkylines(self, left, right):
results = []
l_idx, r_idx = 0, 0
left_height, right_height = 0, 0
while l_idx < len(left) and r_idx < len(right):
point_left = left[l_idx][0]
point_right = right[r_idx][0]
if point_left < point_right:
left_height = left[l_idx][1]
current_x = point_left
current_height = max(left_height, right_height)
l_idx += 1
elif point_left > point_right:
right_height = right[r_idx][1]
current_x = point_right
current_height = max(left_height, right_height)
r_idx += 1
else:
left_height = left[l_idx][1]
right_height = right[r_idx][1]
current_x = point_left
current_height = max(left_height, right_height)
l_idx += 1
r_idx += 1
if not results or results[-1][1] != current_height:
results.append([current_x, current_height])
while l_idx < len(left):
results.append(left[l_idx])
l_idx += 1
while r_idx < len(right):
results.append(right[r_idx])
r_idx += 1
return results
Solve the skyline problem effectively with a divide-and-conquer approach in Python. The solution involves recursively dividing buildings into two halves until the base case, where either zero or one building is left, is reached. For each building set, you compute its skyline representation as pairs of x-coordinates and heights.
- Start by defining a
computeSkylinefunction with recursive calls to handle the left and right halves of the building set. - Introduce a
combineSkylinesfunction to merge two skyline lists into a complete skyline. This function makes use of two pointers for both skyline lists, comparing their x-coordinates to decide from which list the next point will be taken. - Use max logic to ensure the current height reflects the visible skyline profile by comparing heights of current points in both skylines.
- Check for changes in height from the previous point to decide if a point should be added to the results. Points are added if they introduce a change in the skyline's outline.
By merging the skylines of smaller segments iteratively, you achieve an efficient computation of the overall skyline. Handling subproblems ensures that the complexity is manageable, making this approach much more scalable than straightforward comparison methods. The merged results correctly reflect transitions in the skyline height, ensuring an accurate visual representation of the silhouette formed by the input set of buildings.
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