Question

1840. Maximum Building Height

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Description

You want to build n new buildings in a city. The new buildings will be built in a line and are labeled from 1 to n.

However, there are city restrictions on the heights of the new buildings:

• The height of each building must be a non-negative integer.
• The height of the first building must be 0.
• The height difference between any two adjacent buildings cannot exceed 1.

Additionally, there are city restrictions on the maximum height of specific buildings. These restrictions are given as a 2D integer array restrictions where restrictions[i] = [idi, maxHeighti] indicates that building idi must have a height less than or equal to maxHeighti.

It is guaranteed that each building will appear at most once in restrictions, and building 1 will not be in restrictions.

Return _the maximum possible height of the tallest building_.

 

Example 1:

Input: n = 5, restrictions = [[2,1],[4,1]]
Output: 2
Explanation: The green area in the image indicates the maximum allowed height for each building.
We can build the buildings with heights [0,1,2,1,2], and the tallest building has a height of 2.

Example 2:

Input: n = 6, restrictions = []
Output: 5
Explanation: The green area in the image indicates the maximum allowed height for each building.
We can build the buildings with heights [0,1,2,3,4,5], and the tallest building has a height of 5.

Example 3:

Input: n = 10, restrictions = [[5,3],[2,5],[7,4],[10,3]]
Output: 5
Explanation: The green area in the image indicates the maximum allowed height for each building.
We can build the buildings with heights [0,1,2,3,3,4,4,5,4,3], and the tallest building has a height of 5.

 

Constraints:

• 2 <= n <= 109
• 0 <= restrictions.length <= min(n - 1, 105)
• 2 <= idi <= n
• idi is unique.
• 0 <= maxHeighti <= 109

Solutions

Solution 1

class Solution:
  def maxBuilding(self, n: int, restrictions: List[List[int]]) -> int:
    r = restrictions
    r.append([1, 0])
    r.sort()
    if r[-1][0] != n:
      r.append([n, n - 1])
    m = len(r)
    for i in range(1, m):
      r[i][1] = min(r[i][1], r[i - 1][1] + r[i][0] - r[i - 1][0])
    for i in range(m - 2, 0, -1):
      r[i][1] = min(r[i][1], r[i + 1][1] + r[i + 1][0] - r[i][0])
    ans = 0
    for i in range(m - 1):
      t = (r[i][1] + r[i + 1][1] + r[i + 1][0] - r[i][0]) // 2
      ans = max(ans, t)
    return ans

class Solution {
  public int maxBuilding(int n, int[][] restrictions) {
    List<int[]> r = new ArrayList<>();
    r.addAll(Arrays.asList(restrictions));
    r.add(new int[] {1, 0});
    Collections.sort(r, (a, b) -> a[0] - b[0]);
    if (r.get(r.size() - 1)[0] != n) {
      r.add(new int[] {n, n - 1});
    }
    int m = r.size();
    for (int i = 1; i < m; ++i) {
      int[] a = r.get(i - 1), b = r.get(i);
      b[1] = Math.min(b[1], a[1] + b[0] - a[0]);
    }
    for (int i = m - 2; i > 0; --i) {
      int[] a = r.get(i), b = r.get(i + 1);
      a[1] = Math.min(a[1], b[1] + b[0] - a[0]);
    }
    int ans = 0;
    for (int i = 0; i < m - 1; ++i) {
      int[] a = r.get(i), b = r.get(i + 1);
      int t = (a[1] + b[1] + b[0] - a[0]) / 2;
      ans = Math.max(ans, t);
    }
    return ans;
  }
}

class Solution {
public:
  int maxBuilding(int n, vector<vector<int>>& restrictions) {
    auto&& r = restrictions;
    r.push_back({1, 0});
    sort(r.begin(), r.end());
    if (r[r.size() - 1][0] != n) r.push_back({n, n - 1});
    int m = r.size();
    for (int i = 1; i < m; ++i) {
      r[i][1] = min(r[i][1], r[i - 1][1] + r[i][0] - r[i - 1][0]);
    }
    for (int i = m - 2; i > 0; --i) {
      r[i][1] = min(r[i][1], r[i + 1][1] + r[i + 1][0] - r[i][0]);
    }
    int ans = 0;
    for (int i = 0; i < m - 1; ++i) {
      int t = (r[i][1] + r[i + 1][1] + r[i + 1][0] - r[i][0]) / 2;
      ans = max(ans, t);
    }
    return ans;
  }
};

func maxBuilding(n int, restrictions [][]int) (ans int) {
  r := restrictions
  r = append(r, []int{1, 0})
  sort.Slice(r, func(i, j int) bool { return r[i][0] < r[j][0] })
  if r[len(r)-1][0] != n {
    r = append(r, []int{n, n - 1})
  }
  m := len(r)
  for i := 1; i < m; i++ {
    r[i][1] = min(r[i][1], r[i-1][1]+r[i][0]-r[i-1][0])
  }
  for i := m - 2; i > 0; i-- {
    r[i][1] = min(r[i][1], r[i+1][1]+r[i+1][0]-r[i][0])
  }
  for i := 0; i < m-1; i++ {
    t := (r[i][1] + r[i+1][1] + r[i+1][0] - r[i][0]) / 2
    ans = max(ans, t)
  }
  return ans
}