# 36 Valid Sudoku

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.
The Sudoku board could be partially filled, where empty cells are filled with the character '.'.
A partially filled sudoku which is valid.
Note: A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
• A Sudoku board is valid if there exists no duplicates for every row, column, and 3x3 square. That is all I am checking for.
bool scan3x3(vector<vector<char>>& board, int r_start, int c_start) {
unordered_set<int> nums;
for (int i = r_start; i < r_start + 3; i++) {
for (int j = c_start; j < c_start + 3; j++) {
if (board.at(i).at(j) == '.') {
continue;
}
//cout << board.at(i).at(j) << " ";
auto found = nums.find(board.at(i).at(j));
if (found == nums.end()) {
nums.insert(board.at(i).at(j));
}
else {
return false;
}
}
//cout << endl;
}
return true;
}
bool isValidSudoku(vector<vector<char>>& board) {
unordered_set<char> nums;
// scan every row
for (int i = 0; i < board.size(); i++) {
for (int j = 0; j < board.at(i).size(); j++) {
if (board.at(i).at(j) == '.') {
continue;
}
auto found = nums.find(board.at(i).at(j));
if (found == nums.end()) {
nums.insert(board.at(i).at(j));
}
else {
return false;
}
}
nums.clear();
}
// scan every column
for (int i = 0; i < board.size(); i++) {
for (int j = 0; j < board.at(i).size(); j++) {
if (board.at(j).at(i) == '.') {
continue;
}
auto found = nums.find(board.at(j).at(i));
if (found == nums.end()) {
nums.insert(board.at(j).at(i));
}
else {
return false;
}
}
nums.clear();
}
// scan every 3x3
for (int i = 0; i <= 6; i += 3) {
for (int j = 0; j <= 6; j += 3) {
if (scan3x3(board, i, j) == false) {
return false;
}
//pause();
}
}
return true;
}