Description
https://leetcode.com/problems/valid-sudoku/
Determine if a 9×9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:
- Each row must contain the digits
1-9
without repetition. - Each column must contain the digits
1-9
without repetition. - Each of the 9
3x3
sub-boxes of the grid must contain the digits1-9
without 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 digits
1-9
and the character'.'
. - The given board size is always
9x9
.
Accepted
Explanation
To satisfy the following:
- There is no rows with duplicates.
- There is no columns with duplicates.
- There is no sub-boxes with duplicates.
Python Solution
class Solution:
def isValidSudoku(self, board: List[List[str]]) -> bool:
rows = [{} for i in range(9)]
columns = [{} for i in range(9)]
boxes = [{} for i in range(9)]
for i in range(9):
for j in range(9):
num = board[i][j]
if num != '.':
box_index = (i // 3) * 3 + j // 3
rows[i][num] = rows[i].get(num, 0) + 1
columns[j][num] = columns[j].get(num, 0) + 1
boxes[box_index][num] = boxes[box_index].get(num, 0) + 1
if rows[i][num] > 1 or columns[j][num] > 1 or boxes[box_index][num] > 1:
return False
return True
- Time complexity: O(1). Only one iteration over the board with 81 cells.
- Space complexity : O(1).
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