# Valid Sudoku

Published: Sep 26, 2022

Medium Hash Table Matrix

## Introduction

The idea to solve this problem is not difficult. Going over each horizontal, vertical and box to check values are all. The improvement would be how to eliminate loops and run faster.

## Problem Description

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 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.

Constraints:

• `board.length == 9`
• `board[i].length == 9`
• `board[i][j]` is a digit 1-9 or ‘.’.

https://leetcode.com/problems/valid-sudoku/

## Examples

``````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
``````

## Analysis

Prepare two arrays of set for verticals and boxes, and a set for horizontal. The horizontal check is easy. Clear the horizontal set every time row increments. The vertical check is easy as well. A column index is the vertical array’s id. The box check need a bit of consideration. When the current row number modulo 3 is 0, box row id is incremented. The box column id is the same. The box id among 9 is calculated by (box row id * 3 + column / 3). While checking horizontal, vertical, and box numbers, if duplicates are found return False immediately. When all could be checked, the board is valid.

## Solution

``````class ValidSudoku:
def isValidSudoku(self, board: List[List[str]]) -> bool:
vertical = [set() for _ in range(9)]
horizontal = set()
boxes, row_box = [set() for _ in range(9)], -1
for i in range(9):
horizontal.clear()
if i % 3 == 0:
row_box += 1
for j in range(9):
c = board[i][j]
if c == ".":
continue
if c in horizontal:
return False
if c in vertical[j]:
return False
• Time: `O(1)` – double loop is 9 x 9, constant
• Space: `O(1)` – auxiliary array sizes are fixed, constant