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Max Increase to Keep City Skyline in Python
Suppose we have a 2-dimensional array called grid, where each value of grid[i][j] represents the height of a building located there. We can increase the height of any number of buildings, by any amount. Height 0 is considered to be a building as well. At the end, the "skyline" when viewed from all four directions of the grid, must be the same as the skyline of the original grid. Because a city's skyline is the outer contour of the rectangles formed by all the buildings when viewed from a distance. So we have to find the maximum total sum that the height of the buildings can be increased.
So, if the input is like
| 3 | 0 | 8 | 4 |
| 2 | 4 | 5 | 7 |
| 9 | 2 | 3 | 6 |
| 0 | 3 | 1 | 0 |
then the output will be 35, this is because the skyline viewed from top or bottom is: [9, 4, 8, 7], the skyline viewed from left or right is: [8, 7, 9, 3], So the final matrix can be like −
| 8 | 4 | 8 | 7 |
| 7 | 4 | 7 | 7 |
| 9 | 4 | 8 | 7 |
| 3 | 3 | 3 | 3 |
To solve this, we will follow these steps −
max_row_wise := a new list
max_column_wise := a new list
counter := 0
for each i in grid, do
insert maximum of i at the end of max_row_wise
counter := counter + 1
counter := 0, i := 0, j := 0
temp_list := a new list
Do the following infinitely −
insert grid[i,j] into temp_list
i := i + 1
if j is same as size of grid[0] -1 and i>=len(grid), then
insert maximum of temp_list at the end of max_column_wise
come out from the loop
otherwise when i >= size of grid , then
i := 0, j := j + 1
insert maximum of temp_list at the end of max_column_wise
counter := counter + 1
temp_list:= a new list
top_bottom, left_right := max_row_wise,max_column_wise
i, j, value := 0,0,0
Do the following infinitely, do
temp := minimum of [top_bottom[i], left_right[j]]
j := j + 1
if j is same as column length of grid and i is same as row count of grid -1, then
come out from the loop
otherwise when j is same as size of grid column , then
i := i+1
j := 0
return value
Example
Let us see the following implementation to get a better understanding −
class Solution: def maxIncreaseKeepingSkyline(self, grid): max_row_wise = [] max_column_wise = [] counter = 0 for i in grid: max_row_wise.append(max(i)) counter+=1 counter = 0 i = 0 j = 0 temp_list = [] while True: temp_list.append(grid[i][j]) i+=1 if j ==len(grid[0])-1 and i>=len(grid): max_column_wise.append(max(temp_list)) break elif i >= len(grid): i = 0 j = j + 1 max_column_wise.append(max(temp_list)) counter +=1 temp_list=[] top_bottom, left_right = max_row_wise,max_column_wise i, j, value = 0,0,0 while True: temp = min([top_bottom[i], left_right[j]]) value+= abs(grid[i][j] - temp) j+=1 if j == len(grid[0]) and i==len(grid)-1: break elif j == len(grid[0]): i = i+1 j = 0 return value ob = Solution() print(ob.maxIncreaseKeepingSkyline([[3,0,8,4],[2,4,5,7],[9,2,6,3],[0, 3,1,0]]))
Input
[[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
Output
35
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