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:

<strong>Input:</strong> n = 10, restrictions = [[5,3],[2,5],[7,4],[10,3]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> 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

Solution: Two Passes

Trim the max heights based on neighboring max heights.
Two passes: left to right, right to left.

Time complexity: O(n)
Space complexity: O(1)

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

The post 花花酱 LeetCode 1840. Maximum Building Height appeared first on Huahua's Tech Road.