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# Valid Sudoku LeetCode Solution

## Problem – Valid Sudoku

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 `1-9` or `'.'`.

### Valid Sudoku LeetCode Solution in Java

``````public boolean isValidSudoku(char[][] board) {
Set seen = new HashSet();
for (int i=0; i<9; ++i) {
for (int j=0; j<9; ++j) {
if (board[i][j] != '.') {
String b = "(" + board[i][j] + ")";
return false;
}
}
}
return true;
}
``````

### Valid Sudoku LeetCode Solution in C++

``````class Solution
{
public:
bool isValidSudoku(vector<vector<char> > &board)
{
int used1[9][9] = {0}, used2[9][9] = {0}, used3[9][9] = {0};

for(int i = 0; i < board.size(); ++ i)
for(int j = 0; j < board[i].size(); ++ j)
if(board[i][j] != '.')
{
int num = board[i][j] - '0' - 1, k = i / 3 * 3 + j / 3;
if(used1[i][num] || used2[j][num] || used3[k][num])
return false;
used1[i][num] = used2[j][num] = used3[k][num] = 1;
}

return true;
}
};
``````

### Valid Sudoku LeetCode Solution in Python

``````def isValidSudoku(self, board):
return (self.is_row_valid(board) and
self.is_col_valid(board) and
self.is_square_valid(board))

def is_row_valid(self, board):
for row in board:
if not self.is_unit_valid(row):
return False
return True

def is_col_valid(self, board):
for col in zip(*board):
if not self.is_unit_valid(col):
return False
return True

def is_square_valid(self, board):
for i in (0, 3, 6):
for j in (0, 3, 6):
square = [board[x][y] for x in range(i, i + 3) for y in range(j, j + 3)]
if not self.is_unit_valid(square):
return False
return True

def is_unit_valid(self, unit):
unit = [i for i in unit if i != '.']
return len(set(unit)) == len(unit)
``````
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