Valid Sudoku

Search a 2D matrix can be prerequisite for this problem. Use mid/n to get row index,  use mid%n to get column index.

Maybe Spiral Matrix can use the same technique. 


Solution 1:
Row, column and cube can be checked in one loop. When checking ith row, by swaping i and j, can check ith column. Need to memorize the cube index calculation.

Divide the board into 9 3x3. The rowIndex and colIndex are starting indexes of each 3x3. With this information is very easy to iterate through the 3x3 with j incrementing.

class Solution {
    public boolean isValidSudoku(char[][] board) {
        if (board == null || board[0].length == 0) return false;
        
        for (int i = 0; i < board.length; i++) {
                Set<Character> rowSet = new HashSet<>();
                Set<Character> colSet = new HashSet<>();
                Set<Character> cubeSet = new HashSet<>();
            for (int j = 0; j < board[0].length; j++) {
               
                
                // check row
                if (board[i][j] != '.' && !rowSet.add(board[i][j])) return false;
                if (board[j][i] != '.' && !colSet.add(board[j][i])) return false;
                
                int rowIndex = 3 * (i / 3);
                int colIndex = 3 * (i % 3) ;
                if (board[rowIndex + j / 3][colIndex + j % 3] != '.' 

                && !cubeSet.add(board[rowIndex + j / 3][colIndex + j % 3])) return false;

            }
        }
        
        return true;
    } 
}

Solution 2:

Check row, col and cube one by one. It's easier to come up with the cube's indexes. But with much larger time complexity.

Check the solution of Sudoku Solver.