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[LeetCode 36] Valid Sudoku
Problem
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
1-9without repetition. - Each column must contain the digits
1-9without repetition. - Each of the nine
3 x 3sub-boxes of the grid must contain the digits1-9without 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 '.'.
Thoughts
- We have row, col, and square to check uniqueness
- Use set for each
Typescript
function isValidSudoku(board: string[][]): boolean {
let set = new Set()
for(let r = 0; r < board.length; r++) {
for(let c = 0; c < board[r].length; c++){
let val = board[r][c]
if(val === '.') continue;
let rowKey = `r: ${r}, val: ${val}`
let colKey = `c: ${c}, val: ${val}`
// 3*Math.floor(r/3) + Math.floor(c/3) to get values from 0 to 8, identifying
// the 9 large squares, the value in each square must be unique
let sqrKey = `sqr: ${3*Math.floor(r/3) + Math.floor(c/3)}, val: ${val}`
if(set.has(rowKey) || set.has(colKey) || set.has(sqrKey)) return false;
set.add(rowKey)
set.add(colKey)
set.add(sqrKey)
}
}
return true;
};