Question

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?
大致意思：grid 里面的数字代表这现在每个房子的高度。我们需要提高某些建筑物的高度，要求是从四个方向看去，高度是不变得。从上往下看的时候，就是这每一列的最高值，从左往右看的时候是每一行的最高值。

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.

Thought(思路)

Looking this table, we can find thought. You only compare 9, 4, 8, 7, and 8 which one is smaller. We can get the right numbers 8, 4, 8, 7.

#The skyline viewed from top or bottom is: [9, 4, 8, 7]
#The skyline viewed from left or right is: [8, 7, 9, 3]
__|_9__4__8__7__
8 | 8, 4, 8, 7
7 | 7, 4, 7, 7
9 | 9, 4, 8, 7
3 | 3, 3, 3, 3

Code

#the First solution
class Solution(object):
    def maxIncreaseKeepingSkyline(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        gridNew = [[0] * len(grid[0]) for _ in range(len(grid))] 
        top = [max(grid[rows][cols] for rows in range(len(grid))) for cols in range(len(grid[0]))]
        left = [max(grid[rows][cols] for cols in range(len(grid[0]))) for rows in range(len(grid))]
        for row, row_max in enumerate(left):
            for col, col_max in enumerate(top):
                gridNew[row][col] = min(row_max, col_max)
        return sum(gridNew[row][col] - grid[row][col] for row in range(len(left)) for col in range(len(top)))
#the Second solution
class Solution:
    def maxIncreaseKeepingSkyline(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        rows, cols = list(map(max, grid)), list(map(max, zip(*grid)))
        return sum(min(i, j) for i in rows for j in cols) - sum(map(sum, grid))

Complexity

Time complexity: $O(m * n)$ m is row length of grid, n is column length of grid
Space complexity: $O(m+n)$

LeetCode #807. Max Increase to Keep City Skyline python