36. Valid Sudoku¶

Difficulty: Medium

36. Valid Sudoku

Medium

Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

Each row must contain the digits 1-9 without repetition.
Each column must contain the digits 1-9 without repetition.
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 '.'.

Solution¶

class Solution {
    public boolean isValidSudoku(char[][] board) {
        int row = board.length, col = board[0].length;

        for (int i = 0; i < row; i++)
            if (!validRow(board, i))
                return false;

        for (int j = 0; j < col; j++) {
            if (!validCol(board, j))
                return false;
        }

        for (int i = 0; i < row; i += 3) {
            for (int j = 0; j < col; j += 3) {
                if (!validSquare(board, i, j))
                    return false;
            }
        }
        return true;
    }

    public static boolean validRow(char[][] board, int row) {
        boolean vis[] = new boolean[10];
        for(int j  = 0; j < board[0].length; j++) {
            if (board[row][j] == '.') 
                continue;
            if (vis[board[row][j] - '1'] == true)
                return false; 
            vis[board[row][j] - '1'] = true;
        }
        return true;
    }

    public static boolean validCol(char board[][], int col) {
        boolean vis[] = new boolean[10];
        for(int i = 0; i < board.length; i++) {
            if (board[i][col] == '.') 
                continue;
            if (vis[board[i][col] - '1'] == true)
                return false;
            vis[board[i][col] - '1'] = true;
        }
        return true;
    }

    public static boolean validSquare(char board[][], int row, int col) {
        boolean vis[] = new boolean[10];
        for (int i = row; i < row + 3; i++) {
            for (int j =  col; j < col + 3; j++) {
                if (board[i][j] == '.') 
                    continue;
                if (vis[board[i][j] - '1'] == true) 
                    return false;
                vis[board[i][j] - '1'] = true; 
            }
        }
        return true;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here