[Leetcode] Sudoku Solver

Okay, this is a pretty classical backtracking problem. There are some details of writing recursion in python that I can't say I fully understand.

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Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

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I referred to this post to finally get my program running. One thing I did not do is to "return True/False", but handled it like permutation that kind of questions. I do think "return True/False" make sense, but I don't know why I did not think it necessary.

class Solution(object):
    def solveSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        
        self.solve(board, 0, 0)
    
    def solve(self, board, i, j):
        
        def candidates(i, j):
            not_available = []
            for x in range(9):
                if board[i][x] != '.':
                    not_available.append(board[i][x])
                if board[x][j] != '.':
                    not_available.append(board[x][j])
            grid_x, grid_y = i / 3, j / 3
            for x in (0, 1, 2):
                for y in (0, 1, 2):
                    if board[grid_x * 3 + x][grid_y * 3 + y] != '.':
                        not_available.append(board[grid_x * 3 + x][grid_y * 3 + y])
            return [i for i in '123456789' if i not in not_available]
        
        if i == 9: 
            return True
        elif j == 9:
            return self.solve(board, i+1, 0)
        else:            
            if board[i][j] == '.':
                cands = candidates(i, j)
                if not cands:
                    return False
                for cand in cands:
                    board[i][j] = cand
                    if self.solve(board, i, j + 1):
                        return True
                    board[i][j] = '.'
                return False
            else:
                return self.solve(board, i, j+1)