LeetCode - Image Smoother

Given a 2D integer matrix M representing the gray scale of an image, you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.

Example 1:
Input:
[[1,1,1],
 [1,0,1],
 [1,1,1]]
Output:
[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]
Explanation:
For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0
For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0
For the point (1,1): floor(8/9) = floor(0.88888889) = 0

Note:

1.     The value in the given matrix is in the range of [0, 255].
2.     The length and width of the given matrix are in the range of [1, 150].

Source: LeetCode

Solution: There is no trick or such in this problem. It's a straightforward question, all we have to do here is to add all possible cells around a given point.Let's say the give is (x,y) then the possible cells around it will be the following:

  (x-1,y-1)| (x-1,y) |(x-1,y+1)
  (x, y-1) |  {x,y}  |(x-1,y+1)
  (x+1,y-1)| (x+1,y) |(x+1,y+1)

Java Solution:

public int[][] imageSmoother(int[][] M) {
        int row = M.length, col = M[0].length;       
        int[][] result = new int[row][col];  
          
        for(int i=0;i<row;++i){ 

            for(int j=0;j<col;++j){      

                int total = 0, points = 0;

                     for(int k=i-1;k<i+2;++k){ 

                         if(k>-1 && k<row){

                             for(int l=j-1;l<j+2;++l){      

                                  if (l>-1 && l<col){
                                      total+=M[k][l];        
                                      ++points;
                                  }
                              }
                         }
                     }
                result[i][j] = (int) Math.floor(total/(double)points);
            }
        }
        
        return result;
    }