Valid Sudoku – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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

1. Each row must contain the digits 1-9 without repetition.
2. Each column must contain the digits 1-9 without repetition.
3. 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 '.'.

C++ Valid Sudoku LeetCode Solution

Three flags are used to check whether a number appears.
used1: check each row
used2: check each column
used3: check each sub-boxes
class Solution
{
public:
    bool isValidSudoku(vector<vector<char> > &board)
    {
        int used1[9][9] = {0}, used2[9][9] = {0}, used3[9][9] = {0};
        
        for(int i = 0; i < board.size(); ++ i)
            for(int j = 0; j < board[i].size(); ++ j)
                if(board[i][j] != '.')
                {
                    int num = board[i][j] - '0' - 1, k = i / 3 * 3 + j / 3;
                    if(used1[i][num] || used2[j][num] || used3[k][num])
                        return false;
                    used1[i][num] = used2[j][num] = used3[k][num] = 1;
                }
        
        return true;
    }
};

Java Valid Sudoku LeetCode Solution

public boolean isValidSudoku(char[][] board) {
    Set seen = new HashSet();
    for (int i=0; i<9; ++i) {
        for (int j=0; j<9; ++j) {
            char number = board[i][j];
            if (number != '.')
                if (!seen.add(number + " in row " + i) ||
                    !seen.add(number + " in column " + j) ||
                    !seen.add(number + " in block " + i/3 + "-" + j/3))
                    return false;
        }
    }
    return true;
}
 

Python 3 Valid Sudoku LeetCode Solution

def isValidSudoku(self, board):
    return (self.is_row_valid(board) and
            self.is_col_valid(board) and
            self.is_square_valid(board))

def is_row_valid(self, board):
    for row in board:
        if not self.is_unit_valid(row):
            return False
    return True

def is_col_valid(self, board):
    for col in zip(*board):
        if not self.is_unit_valid(col):
            return False
    return True
    
def is_square_valid(self, board):
    for i in (0, 3, 6):
        for j in (0, 3, 6):
            square = [board[x][y] for x in range(i, i + 3) for y in range(j, j + 3)]
            if not self.is_unit_valid(square):
                return False
    return True
    
def is_unit_valid(self, unit):
    unit = [i for i in unit if i != '.']
    return len(set(unit)) == len(unit)

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