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MEDIUM

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

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

Explanation:
The board is valid according to the rules above.

---

Constraints

• board.length == 9
• board[i].length == 9
• board[i][j] is a digit 1-9 or '.'.

---

Solution: Hash Table

• Time Complexity: O(1) (since the board size is fixed)
• Space Complexity: O(1)

C++

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        vector<unordered_set<int>> r(9), c(9), b(9);
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char v = board[i][j];
                if (v == '.') continue;
                if (r[i].count(v) || c[j].count(v) || b[i/3*3 + j/3].count(v)) return false;
                r[i].insert(v);
                c[j].insert(v);
                b[i/3*3 + j/3].insert(v);
            }
        }
        return true;
    }
};