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.
In 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.
My approach:
class Solution {
public int maxIncreaseKeepingSkyline(int[][] grid) {
int nrows = grid.length;
int ncols = grid[0].length;
int [] leftRig = new int[nrows];
int [] topBot = new int[ncols];
//Filling topBot
for( int i = 0; i < ncols; i++ )
{
int max = Integer.MIN_VALUE;
for( int j = 0; j < nrows; j++ )
{
if( max <= grid[j][i] )
max = grid[j][i];
}
topBot[i] = max;
}
//Filling leftRig
for( int k = 0; k < nrows; k++ )
{
int max = Integer.MIN_VALUE;
for( int l = 0; l < ncols; l++ )
{
if( max <= grid[k][l] )
max = grid[k][l];
}
leftRig[k] = max;
}
int count = 0;
int min = 0;
//Enumerating the minimum height to be added
for( int i = 0; i < nrows; i++ )
{
for( int j = 0; j < ncols; j++ )
{
min = Math.min(leftRig[i],topBot[j]);
count += min - grid[i][j];
}
}
return count;
}
}
Time complexity: O(n^2)
Space complexity: O(n)
Time complexity: O(n) Space complexity: O(n)
I have the following questions regarding the above code snippets:
1) How can I improve the time and space complexity of my code?
2) Is there a better way(lesser lines of code, better data structures) that can be used to improve the code?
3) Is there any better approach to solve this question?
O(n)
? Because the algorithm works on a given amount of buildings, and you simply iterate over each building a couple of times? \$\endgroup\$n
represent in your problem? I would say, the amount of buildings. Each nested loop iterates all buildings. So I would say you do around3n
operations, which is still linear w.r.t. the input. If I am not mistaken! \$\endgroup\$