From given row(x),column(y), find 3x3 subarray of a 2D 9x9 array

Question

So i'm trying to get all the possible entries that can be put into a Sudoku single square. I have a 9x9 2D array, which is further divided into 3x3 subarrays. I want to write a method that takes a row & column combination in its parameters and returns all possible entries that can be made at that specific position. The first 2 for-loops of my method takes all the already existing non-zero values in the entire row & entire column that is specified and stores them in an array (alreadyInUse),this will at a later stage be used to figure out what numbers are not already used. The third for-loop should, using the row,column combination, find the specific subarray and add its entries to the alreadyInUse-array.

Is there any way to find the row,column of the subarray using given row,column of the 2D array?

    // Method for calculating all possibilities at specific position
public int[] getPossibilities(int col, int row){
    int [] possibilities;
    int [] alreadyInUse = null;
    int currentIndex = 0;
    if(sudoku[row][col] != 0){
        return  new int[]{sudoku[col][row]};
    }
    else{
        alreadyInUse = new int[26];
        //Go into Row x and store all available numbers in an alreadyInUse
        for(int i=0; i<sudoku.length; i++){
            if(sudoku[row][i] !=0){
                alreadyInUse[currentIndex] = sudoku[row][i];
                currentIndex++;
            }
        }
        for(int j=0; j<sudoku.length; j++){
            if(sudoku[j][col] !=0){
                alreadyInUse[currentIndex] = sudoku[j][col];
                currentIndex++;
            }
        }
        for(int k=...???

    }
        return possibilities;
} 

Answer 1

You can use modulus to filter out the sub array. For example, one way to do it would be to use the expression n - (n % 3) . For example, if row is column 8 (the last column in a 0 indexed array), this expression will return 6. It will also return 6 for column 6, but it will return 3 for column 5.

Then, once you have the top left cell, you can loop through all 9 cells using a nested loop, three at a time.

Here's relevant code:

int x_left = (row - (row % 3));
int y_top = (col - (col % 3));
for(int i=0; i<3; i++) {
  for(int j=0; j<3; j++) {
     if(sudoku[i + x_left][j + y_top] != 0) {
       alreadyInUse[currentIndex] = sudoku[i + x_left][j + y_top];
       currentIndex++;
     }
  }
}