Leetcode - Valid Sudoku Solution

Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

1. Each row must contain the digits 1-9 without repetition.
2. Each column must contain the digits 1-9 without repetition.
3. Each of the nine 3 x 3 sub-boxes of the grid must contain the digits 1-9 without repetition.

Note:

• A Sudoku board (partially filled) could be valid but is not necessarily solvable.
• Only the filled cells need to be validated according to the mentioned rules.

Example 1:

Input: board = 
[["5","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
Output: true

Example 2:

Input: board = 
[["8","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Constraints:

• board.length == 9
• board[i].length == 9
• board[i][j] is a digit or '.'.

Solution in Python

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        for row in board:
            c = set()
            for n in row:
                if n!= "." and n in c:
                    return False
                else:
                    c.add(n) 

        for i in range(9):
            c = set()
            for j in range(9):
                n = board[j][i]
                if n != "." and n in c:
                    return False
                else:
                    c.add(n)

        p = [[0,1,2], [3,4,5], [6,7,8]]
        for x in p:
            for y in p:
                box = product(x,y)
                c = set()
                for i, j in box:
                    n = board[j][i]
                    if n != "." and n in c:
                        return False
                    else:
                        c.add(n)
        return True