[leetcode] 807. Max Increase to Keep City Skyline @ python

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原题

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.

解法

代码

首先计算出每列的最大值和每行的最大值, 然后构建new_grid, 先将每列的值都调成每列的最大值, 然后遍历new_grid, 遇到该行的最大值小于当前值时, 改为该行的最大值即可. 最后求两个grid的和的差值.

Time: O(n)
Space: O(n)

class Solution:
    def maxIncreaseKeepingSkyline(self, grid: 'List[List[int]]') -> 'int':
        row, col = len(grid), len(grid[0])
        left = []
        for row_list in grid:
            left.append(max(row_list))
            
        top = []        
        for c in range(col):
            column = [grid[r][c] for r in range(row)]
            top.append(max(column))
        # get the new grid
        new_grid = copy.deepcopy(grid)
        for c in range(col):
            for r in range(row):
                new_grid[r][c] = top[c]
        
        for r in range(row):
            for c in range(col):
                new_grid[r][c] = min(new_grid[r][c], left[r])
                
        sum_of_grid =  sum([sum(row) for row in grid])
        sum_of_new_grid = sum([sum(row) for row in new_grid])
        return sum_of_new_grid - sum_of_grid