36_Valid Sudoku

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

1. Each row must contain the digits 1-9without repetition.

2. Each column must contain the digits 1-9without repetition.

3. Each of the 93x3sub-boxes of the grid must contain the digits 1-9without repetition.

A partially filled sudoku which is valid.

The Sudoku board could be partially filled, where empty cells are filled with the character'.'.

Example 1:

Input:
[
  ["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:
[
  ["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.

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.

• The given board contain only digits1-9and the character'.'.

• The given board size is always9x9

  .

Solution 1:

Idea:

• Loop over each row, column and sub grid to check if duplicates exist

Time Complexity: $O(n^2)$

Space Complexity: $O(n)$

def isValidSudoku(self, board):
    """
    :type board: List[List[str]]
    :rtype: bool
    """
    # define a function to check if a list is valid
    def isValidList(l) :
        nums = [l[i] for i in range(len(l)) if l[i] != '.']
        return len(nums) == len(set(nums))

    # check rows
    for i in range(9) :
        if not isValidList(board[i]):
            return False

    # check columns
    for j in range(9) :
        l = [board[i][j] for i in range(9)]
        if not isValidList(l) :
            return False

    # check small grids 
    for m in range(3) :
        for n in range(3) :
            l = [board[i][j] for i in range(3*m, 3*m+3) for j in range(3*n, 3*n+3)]
            if not isValidList(l) :
                return False

    return True

Solution 2:

Loop once instead of 3 times as in solution 1 sacrificing space complexity as $O(n^2)$

def isValidSudoku(self, board):
    """
    :type board: List[List[str]]
    :rtype: bool
    """
    seen = sum(([(c, i), (j, c), (i/3, j/3, c)] 
                for i, row in enumerate(board)
                for j, c in enumerate(row)
                if c != '.'), [])
    return len(seen) == len(set(seen))