Question

link

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.

Stats

Frequency 2
Difficulty 2
Adjusted Difficulty 2
Time to use ----------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Solution

This is not a difficult problem.

Make use of three for-loops and nine arrays of length 9 (for each loop) to mark the status, then do DFS search.

However, I also found a very concise solution. Read below.

My code

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        if (board == null || board.length == 0) {
            return false;
        }
        int N = board.length;
        for (int i = 0; i < N; i++) {
            boolean[] foo = new boolean[N];
            // validate each row
            for (int j = 0; j < N; j++) {
                if (board[i][j] != '.') {
                    if (foo[board[i][j] - '1']) {
                        return false;
                    }
                    foo[board[i][j] - '1'] = true;
                }
            }
            foo = new boolean[N];
            // validate each column
            for (int j = 0; j < N; j++) {
                if (board[j][i] != '.') {
                    if (foo[board[j][i] - '1']) {
                        return false;
                    }
                    foo[board[j][i] - '1'] = true;
                }
            }
        }
        for (int a = 0; a < 3; a++) {
            for (int b = 0; b < 3; b++) {
                boolean[] foo = new boolean[N];
                for (int c = 0; c < 3; c++) {
                    for (int d = 0; d < 3; d++) {
                        if (board[a * 3 + c][b * 3 + d] != '.') {
                            if (foo[board[a * 3 + c][b * 3 + d] - '1']) {
                                return false;
                            }
                            foo[board[a * 3 + c][b * 3 + d] - '1'] = true;
                        }
                    }
                }
            }
        }
        return true;
    }
}

The following solution is from this blog. It’s a very clever and surprisingly concise code.

public boolean isValidSudoku(char[][] board) {
    boolean[][] rows = new boolean[9][9];
    boolean[][] cols = new boolean[9][9];
    boolean[][] blocks = new boolean[9][9];
    for (int i = 0; i < 9; ++i) {
        for (int j = 0; j < 9; ++j) {
            int c = board[i][j] - '1';
            if (board[i][j] == '.') continue;
            if (rows[i][c] || cols[j][c] || blocks[i - i % 3 + j / 3][c])
                return false;
            rows[i][c] = cols[j][c] = blocks[i - i % 3 + j / 3][c] = true;
        }
    }
    return true;
}