Max Increase to Keep City Skyline in Java 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 &lt; grid.length = grid[0].length &lt;= 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 &lt; ncols; i++ )
        {
               int max = Integer.MIN_VALUE;    

                for( int j = 0; j &lt; nrows; j++ )
               {
                   if( max &lt;= grid[j][i] )
                       max = grid[j][i];
               }
            topBot[i] = max;
        }

        //Filling leftRig
        for( int k = 0; k &lt; nrows; k++ )
        {
               int max = Integer.MIN_VALUE;    

                for( int l = 0; l &lt; ncols; l++ )
               {
                   if( max &lt;= 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 &lt; nrows; i++ )
        {       
          for( int j = 0; j &lt; 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?

Reference
Here:

for( int i = 0; i &lt; ncols; i++ )
{
    int max = Integer.MIN_VALUE;

    for( int j = 0; j &lt; nrows; j++ )
    {
        if( max &lt;= grid[j][i] )
            max = grid[j][i];
    }
    topBot[i] = max;
}


You don't need an extra variable max as a reference. You can simply use topBot[i] as a reference instead (the elements of an int array will be initialized to the default value 0 upon the array's creation, and since the height of a building cannot be less than 0, 0 is an adequate initial value). Also, it would suffice to check if(max &lt; grid[j][i]) (i.e. using &lt; instead of &lt;=), because you don't need to assign the variable if the two values are equal. In fact, there's an even more concise way to write this:

topBot[i] = Math.max(topBot[i], grid[j][i]);


And with the local variable max eliminated, you can fill both topBot and leftRig in one loop:

for (int i = 0; i &lt; ncols; i++) {
    for (int j = 0; j &lt; nrows; j++) {
        topBot[i] = Math.max(topBot[i], grid[j][i]);
        leftRig[j] = Math.max(leftRig[j], grid[j][i]);
    }
}

In your final loop, min is not needed outside the inner for loop, so it would suffice to declare it inside this inner loop. Reducing the scope of variables to the smallest extent necessary makes code easier to read, because, in this case, it would make it instantly obvious that the variable is not needed outside the loop.