36 Valid Sudoku
Determine if a Sudoku is valid, according to:Sudoku Puzzles - The Rules.
The Sudoku board could be partially filled, where empty cells are filled with the character'.'
.

A partially filled sudoku which is valid.
Note: A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated. (A completely empty board is a valid square).
The Idea: A board is valid if all the rows, columns, and squares contain no duplicate integers.
Complexity: O(n^2) time and O(n) space with n is the len(sqrt(board))
class Solution(object):
def isValidSudoku(self, board):
"""
:type board: List[List[str]]
:rtype: bool
"""
dim = len(board)
if not board or dim == 0 or \
len(board[0]) != dim:
return False
# rule 1: no duplicates in rows
for row in board:
row_elms = [elm for elm in row if elm != '.']
if len(set(row_elms)) != len(row_elms):
return False
# rule 2: no duplicates in cols
for i in range(dim):
col_elms = [board[j][i] for j in range(dim) if board[j][i] != '.']
if len(set(col_elms)) != len(col_elms):
return False
# rule 3: no duplicates in each square grid
square = int(dim/3)
for row_mult in range(square):
for col_mult in range(square):
square_elm = []
for i in range(row_mult*square, (row_mult*square) + square):
for j in range(col_mult*square, (col_mult*square) + square):
if board[i][j] != '.':
square_elm.append(board[i][j])
if len(set(square_elm)) != len(square_elm):
return False
return True
Discussion
If the board is not necessarily 9x9, and if we wanted to generalize this algorithm. Here are some of the changes that need to be made.
The board must be a perfect square. e.g.
0^2, 1^1, 2^2 ... , n^n
square
variable should then becomemath.sqrt(dim)
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