How to validate a Sudoku board

The Problem

With this article, I will be covering the valid sudoku Leetcode problem.

Leetcode describes the problem with the following:

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.

Leetcode ranks this problem as a medium. I think that is an appropriate rating. The solution is feasible but does require some out-of-the-box reasoning.

The Solution

    def isValidSudoku(self, board):
        seen = set()
        for i in range(9):
            for j in range(9):
                number = board[i][j]
                if number != ".":
                    row = str(number) + "row" + str(i)
                    column = str(number) + "column" + str(j)
                    block = str(number) + "block" + str(i / 3) + "/" + str(j/3)
                    if row in seen or column in seen or block in seen:
                        return False
                    seen.add(row)
                    seen.add(column)
                    seen.add(block)
        return True

Solution Description

1. seen = set(): Creates an empty set named seen to keep track of numbers that have already appeared in rows, columns, and 3x3 blocks.

2. Two nested loops for i in range(9) and for j in range(9) iterate through each cell of the 9x9 board.

3. number = board[i][j]: Stores the current cell's value in the variable number.

4. if number != ".":: Checks if the cell is not empty (a dot indicates an empty cell in this context).

5. row = str(number) + "row" + str(i): Creates a string like 1row0 to indicate that the number 1 is in row 0.

6. column = str(number) + "column" + str(j): Creates a string like 1column0 to indicate that the number 1 is in column 0.

7. block = str(number) + "block" + str(i / 3) + "/" + str(j/3): Creates a string like 1block0/0 to indicate that the number 1 is in the top-left 3x3 block.

8. if row in seen or column in seen or block in seen:: Checks if any of these strings already exist in the seen set. If so, the board is not valid and the function returns False.

9. seen.add(row), seen.add(column), seen.add(block): Adds the newly created strings to the seen set so that they can be checked against future cells.

10. If all cells are valid, the function returns True, indicating a valid Sudoku board.

Big O Analysis

Assuming n is the side length. We have a double for-loop so that is at least O(n^2). Inside the nested loops, the operations (like adding to a set, creating strings, and checking for membership in a set) are O(1) operations. Therefore, the overall runtime is O(n^2).

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Originally published at https://blog.seancoughlin.me.