Algorithm to determine if a sudoku puzzle is valid
Determine if a 9×9 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 9 3×3 sub-boxes of the grid must contain the digits 1-9 without repetition.
A partially filled sudoku which is valid.
The Sudoku board could be partially filled, where empty cells are filled with the character ‘.’.
Example 1:
Input:
[
[“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:
[
[“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 3×3 sub-box, it is invalid.
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.
- The given board contain only digits 1-9 and the character ‘.’.
- The given board size is always 9×9.
Posted from https://forum.golibrary.co/question/algorithm-determine-sudoku-puzzle-valid/
Here is another cryptic but short solution in javascript
var isValidSudoku = function(board) {
for (let i = 0; i < 9; i++) {
let row = new Set(), col = new Set(), sqr = new Set();
for (let j = 0; j < 9; j++) {
let rowc = board[i][j];
let colc = board[j][i];
let sqrc = board[Math.floor(i / 3) * 3 + Math.floor(j / 3)][(i % 3) * 3 + j % 3];
if (row.has(rowc) || col.has(colc) || sqr.has(sqrc)) return false;
if (rowc !== ".") row.add(rowc);
if (colc !== ".") col.add(colc);
if (sqrc !== ".") sqr.add(sqrc);
}
}
return true;
};
