# LeetCode 807. Max Increase to Keep City Skyline

In a 2 dimensional array `grid`, each value `grid[i][j]` represents the height of a building located there. We are allowed to increase the height of any number of buildings, by any amount (the amounts can be different for different buildings). Height 0 is considered to be a building as well.

At the end, the "skyline" when viewed from all four directions of the grid, i.e. top, bottom, left, and right, must be the same as the skyline of the original grid. A city's skyline is the outer contour of the rectangles formed by all the buildings when viewed from a distance. See the following example.

What is the maximum total sum that the height of the buildings can be increased?

```Example:
Input: grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
Output: 35
Explanation:
The grid is:
[ [3, 0, 8, 4],
[2, 4, 5, 7],
[9, 2, 6, 3],
[0, 3, 1, 0] ]

The skyline viewed from top or bottom is: [9, 4, 8, 7]
The skyline viewed from left or right is: [8, 7, 9, 3]

The grid after increasing the height of buildings without affecting skylines is:

gridNew = [ [8, 4, 8, 7],
[7, 4, 7, 7],
[9, 4, 8, 7],
[3, 3, 3, 3] ]```

Notes:

• `1 < grid.length = grid.length <= 50`.
• All heights `grid[i][j]` are in the range `[0, 100]`.
• All buildings in `grid[i][j]` occupy the entire grid cell: that is, they are a `1 x 1 x grid[i][j]` rectangular prism.

```示例：

输出： 35
说明：

[[3,0,8,4]，
[2,4,5,7]，
[9,2,6,3]，
[ 0,3,1,0 ]] 查看的天际线从顶部或底部是：[ 9,4,8​​,7 ]从左或右看的天际线是：[ 8,7,9,3 ]

gridNew = [[ 8,4,8,7]，
[7,4,7,7]，
[9,4,8​​,7]，
[3,3,3,3]]```

• `1 < grid.length = grid.length <= 50`.
• 所有高度`grid[i][j]`都在范围内`[0, 100]`
• 所有建筑物都`grid[i][j]`占据整个网格单元：也就是说，它们是一个`1 x 1 x grid[i][j]`矩形棱柱。

```class Solution {
public:
int maxIncreaseKeepingSkyline(vectorint>>& grid) {
int m = grid.size(), n = grid.size(), res = 0;
vector<int> row(m, 0), col(n, 0);
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
row[i] = max(row[i], grid[i][j]);
col[j] = max(col[j], grid[i][j]);
}
}
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
res += min(row[i] - grid[i][j], col[j] - grid[i][j]);
}
}
return res;
}
};```