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:
- 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 digits1-9
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 digit1-9
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)
Array-1180
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