# Ac solution code

• The rules of Sudoku game can be found here:
Grids: 9 x 9

``````1) 1-9 only occurs once in each ROW
2) 1-9 only occurs once in each COLUMN
3) 1-9 only occurs once in each 3 x 3 square (9 squares totally with splitter of length 3)
``````

One pass solution: Special thanks for Lorraine921's one pass idea.

``````public boolean isValidSudoku(char[][] board) {
for (int i= 0; i < board.length; i++) {
Set<Character>row = new HashSet<Character>();// Check row
Set<Character>col = new HashSet<Character>();// Check column
Set<Character>square = new HashSet<Character>();// Check sub square
for (int j = 0; j < board[0].length; j++) {
return false;
if (board[j][i] != '.' && !col.add(board[j][i]))
return false;
int rowIndex = 3 * (i/3), colIndex = 3 * (i % 3);
if (board[rowIndex + j/3][colIndex + j % 3] != '.' && !square.add(board[rowIndex + j/3][colIndex + j % 3]))
return false;
}
}
return true;
}
``````

Three passes solution:

``````public boolean isValidSudoku(char[][] board) {
boolean visited[] = new boolean[9];
for (int i = 0; i < board.length; i++) {//Row
Arrays.fill(visited, false);
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] != '.') {
if (visited[board[i][j] - '0' - 1])
return false;
visited[board[i][j] - '0' - 1] = true;
}
}
}
for (int i = 0; i < board[0].length; i++) {//Column
Arrays.fill(visited, false);
for (int j = 0; j < board.length; j++) {
if (board[j][i] != '.') {
if (visited[board[j][i] - '0' - 1])
return false;
visited[board[j][i] - '0' - 1] = true;
}
}
}
for (int i = 0; i < board.length; i+=3) {//Sub square
for (int j = 0; j < board[0].length; j+=3) {
Arrays.fill(visited, false);
for (int k = 0; k < 9; k++) {
if (board[i+k/3][j+k%3] != '.') {
if (visited[board[i+k/3][j+k%3] - '0' - 1])
return false;
visited[board[i+k/3][j+k%3] - '0' - 1] = true;
}
}
}
}
return true;
}
``````

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.