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# 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[0].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.

#### 解题要点：

遍历二维数组，找出从左至右，从上到下的最高建筑一维数组。然后再遍历一次，基于这个一维数组判断出哪些建筑还有上升空间，并加盖到不超过最高建筑，将这个升高的值加入最终返回值中。

1ms：

```java
class Solution {
    public int maxIncreaseKeepingSkyline(int[][] grid) {
        int res = 0;
        int[] topdown = new int[grid.length];
        int[] leftright = new int[grid.length];

        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid.length; j++){
                leftright[i] = grid[i][j] > leftright[i] ? grid[i][j] : leftright[i];
                topdown[i] = grid[j][i] > topdown[i] ? grid[j][i] : topdown[i];
            }
        }
        
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid.length; j++){
                res += Math.min(leftright[i], topdown[j]) - grid[i][j];
            }
        }
        
        return res;
    }
}
```

2ms：

```java
class Solution {
    public int maxIncreaseKeepingSkyline(int[][] grid) {
        int res = 0;
        int[] topdown = new int[grid.length];
        int[] leftright = new int[grid.length];

        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid.length; j++){
                leftright[i] = grid[i][j] > leftright[i] ? grid[i][j] : leftright[i];
                topdown[i] = grid[j][i] > topdown[i] ? grid[j][i] : topdown[i];
            }
        }
        
        for(int w = 0; w < grid.length; w++){
            for(int e = 0; e < grid.length; e++){
                while(grid[w][e] < leftright[w] && grid[w][e] < topdown[e]){
                    grid[w][e]++;
                    res++;
                }
            }
        }
        
        return res;
    }
}
```