[LeetCode] 36. Valid Sudoku_Medium tag: Array

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 '.'.

 

Idea:用两个for loop 去检测每一个点，然后看该行，该列和所在的3 *3的小正方形中是否有相同的数，如果有表明invalid，最后两个for loop结束，返回True。

1. brute force 

T: O(n * n * n)  n = 9

S: O(1)

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        for i in range(9):
            for j in range(9):
                if board[i][j] == '.':
                    continue
                if not self.isValid(board, i, j):
                    return False
        return True
    
    
    def isValid(self, board, i, j):
        c = board[i][j]
        for k in range(9):
            row = (i //3) * 3 + k //3
            col = (j //3) * 3 + k % 3
            if (k != j and c == board[i][k]) or (k != i and c == board[k][j]) or row != i and col != j and c == board[row][col]:
                return False
        return True

 

 

我们利用3个9 * 9的array去记录，

rowFlag[row][num] 的左下标row表示的是行数，右下标num表示的是（“实际数字” - 1），数值rowFlag[row][num]如果为False表示目前该row行没有出现过num + 1 这个数字，如果为True表明出现过num + 1这个数字。

同理，colFlag[col][num] 的左下标col表示的是列数，右下标num表示的是（“实际数字” - 1），数值colFlag[col][num]如果为False表示目前该col列没有出现过num + 1 这个数字，如果为True表明出现过num + 1这个数字。

最后，cellFlag[cellIndex][num]的左下标cellIndex表示的是哪个小正方形的index, 如下图所示，图是来自Leetcode这道题的solution，右下标num表示的是（“实际数字” - 1），数值cellFlag[cellIndex][num]如果为False表示目前该cellIndex的小正方形没有出现过num + 1 这个数字，如果为True表明出现过num + 1这个数字。

cellIndex = (row //3) * 3 + col //3          , 可以记住，或者自己推一下就可以了，可以带入row 和col进去试试看就明白了。

 

 n = 9 in this case

T: O(n * n)

S: O(n * n)

 

Code

class Solution:
    def isValidSudoku(self, board: List[list[str]]) -> bool:
        rowFlag = [[False] * 9 for _ in range(9)]
        colFlag =  [[False] * 9 for _ in range(9)]
        cellFlag =  [[False] * 9 for _ in range(9)]
        for i in range(9):
            for j in range(9):
                if board[i][j] == '.':
                    continue
                num = int(board[i][j]) - 1
                cellIndex = (i//3) * 3 + j //3
                if rowFlag[i][num] or colFlag[j][num] or cellFlag[cellIndex][num]:
                    return False
                rowFlag[i][num] = True
                colFlag[j][num] = True
                cellFlag[cellIndex][num] = True
        return True