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.
bool isValidSudoku(vector<vector<char>>& board) { // time: O(n^2); space: O(n^2)
if (board.empty() || board[0].empty() || board.size() != board[0].size()) return false;
int n = board.size();
vector<vector<bool> > row(n, vector<bool>(n, false)), col(n, vector<bool>(n, false)), box(n, vector<bool>(n, false));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (board[i][j] == '.') continue;
int k = board[i][j] - '1';
if (row[i][k] || col[j][k] || box[3 * (i / 3) + (j / 3)][k]) return false;
row[i][k] = col[j][k] = box[3 * (i / 3) + (j / 3)][k] = true;
}
}
return true;
}
// Optimized Space
bool isValidSudoku(vector<vector<char>>& board) { // time: O(n^2); space: O(n)
if (board.empty() || board[0].empty() || board.size() != board[0].size()) return false;
int n = board.size();
vector<short> row(n, 0), col(n, 0), box(n, 0);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (board[i][j] == '.') continue;
int mask = 1 << (board[i][j] - '1');
int k = 3 * (i / 3) + (j / 3);
if (row[i] & mask || col[j] & mask || box[k] & mask) return false;
row[i] |= mask;
col[j] |= mask;
box[k] |= mask;
}
}
return true;
}