The Skyline Problem LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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]]
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.

The Skyline Problem Solutions

Time: O(nlogn)
Space: O(n)

C++

 class Solution {
 public:
  vector<vector<int>> getSkyline(const vector<vector<int>>& buildings) {
    const int n = buildings.size();
    if (n == 0)
      return {};
    if (n == 1) {
      const int left = buildings[0][0];
      const int right = buildings[0][1];
      const int height = buildings[0][2];
      return {{left, height}, {right, 0}};
    }

    const auto left = getSkyline({begin(buildings), begin(buildings) + n / 2});
    const auto right = getSkyline({begin(buildings) + n / 2, end(buildings)});
    return merge(left, right);
  }

 private:
  vector<vector<int>> merge(const vector<vector<int>>& left,
                            const vector<vector<int>>& right) {
    vector<vector<int>> ans;
    int i = 0;  // left's index
    int j = 0;  // right's index
    int leftY = 0;
    int rightY = 0;

    while (i < left.size() && j < right.size())
      // choose the point with smaller x
      if (left[i][0] < right[j][0]) {
        leftY = left[i][1];  // update the ongoing leftY
        addPoint(ans, left[i][0], max(left[i++][1], rightY));
      } else {
        rightY = right[j][1];  // update the ongoing rightY
        addPoint(ans, right[j][0], max(right[j++][1], leftY));
      }

    while (i < left.size())
      addPoint(ans, left[i][0], left[i++][1]);

    while (j < right.size())
      addPoint(ans, right[j][0], right[j++][1]);

    return ans;
  }

  void addPoint(vector<vector<int>>& ans, int x, int y) {
    if (!ans.empty() && ans.back()[0] == x) {
      ans.back()[1] = y;
      return;
    }
    if (!ans.empty() && ans.back()[1] == y)
      return;
    ans.push_back({x, y});
  }
};

Java

 class Solution {
  public List<List<Integer>> getSkyline(int[][] buildings) {
    final int n = buildings.length;
    if (n == 0)
      return new ArrayList<>();
    if (n == 1) {
      final int left = buildings[0][0];
      final int right = buildings[0][1];
      final int height = buildings[0][2];
      List<List<Integer>> ans = new ArrayList<>();
      ans.add(new ArrayList<>(Arrays.asList(left, height)));
      ans.add(new ArrayList<>(Arrays.asList(right, 0)));
      return ans;
    }

    var leftSkyline = getSkyline(Arrays.copyOfRange(buildings, 0, n / 2));
    var rightSkyline = getSkyline(Arrays.copyOfRange(buildings, n / 2, n));
    return merge(leftSkyline, rightSkyline);
  }

  private List<List<Integer>> merge(List<List<Integer>> left, List<List<Integer>> right) {
    List<List<Integer>> ans = new ArrayList<>();
    int i = 0; // left's index
    int j = 0; // right's index
    int leftY = 0;
    int rightY = 0;

    while (i < left.size() && j < right.size())
      // choose the point with smaller x
      if (left.get(i).get(0) < right.get(j).get(0)) {
        leftY = left.get(i).get(1); // update the ongoing leftY
        addPoint(ans, left.get(i).get(0), Math.max(left.get(i++).get(1), rightY));
      } else {
        rightY = right.get(j).get(1); // update the ongoing rightY
        addPoint(ans, right.get(j).get(0), Math.max(right.get(j++).get(1), leftY));
      }

    while (i < left.size())
      addPoint(ans, left.get(i).get(0), left.get(i++).get(1));

    while (j < right.size())
      addPoint(ans, right.get(j).get(0), right.get(j++).get(1));

    return ans;
  }

  private void addPoint(List<List<Integer>> ans, int x, int y) {
    if (!ans.isEmpty() && ans.get(ans.size() - 1).get(0) == x) {
      ans.get(ans.size() - 1).set(1, y);
      return;
    }
    if (!ans.isEmpty() && ans.get(ans.size() - 1).get(1) == y)
      return;
    ans.add(new ArrayList<>(Arrays.asList(x, y)));
  }
}

Python

  class Solution:
  def getSkyline(self, buildings: List[List[int]]) -> List[List[int]]:
    n = len(buildings)
    if n == 0:
      return []
    if n == 1:
      left, right, height = buildings[0]
      return [[left, height], [right, 0]]

    left = self.getSkyline(buildings[:n // 2])
    right = self.getSkyline(buildings[n // 2:])
    return self._merge(left, right)

  def _merge(self, left: List[List[int]], right: List[List[int]]) -> List[List[int]]:
    ans = []
    i = 0  # left's index
    j = 0  # right's index
    leftY = 0
    rightY = 0

    while i < len(left) and j < len(right):
      # choose the powith smaller x
      if left[i][0] < right[j][0]:
        leftY = left[i][1]  # update the ongoing leftY
        self._addPoint(ans, left[i][0], max(left[i][1], rightY))
        i += 1
      else:
        rightY = right[j][1]  # update the ongoing rightY
        self._addPoint(ans, right[j][0], max(right[j][1], leftY))
        j += 1

    while i < len(left):
      self._addPoint(ans, left[i][0], left[i][1])
      i += 1

    while j < len(right):
      self._addPoint(ans, right[j][0], right[j][1])
      j += 1

    return ans

  def _addPoint(self, ans: List[List[int]], x: int, y: int) -> None:
    if ans and ans[-1][0] == x:
      ans[-1][1] = y
      return
    if ans and ans[-1][1] == y:
      return
    ans.append([x, y])

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