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Sudoku Solver

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Sudoku Solver

Description

https://leetcode.com/problems/sudoku-solver/description/

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

A sudoku solution must satisfy all of the following rules:

  1. Each of the digits 1-9 must occur exactly once in each row.
  2. Each of the digits 1-9 must occur exactly once in each column.
  3. Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.

Empty cells are indicated by the character '.'.

Solution

This is a brutal-force implementation of Sudoku algorithm, The solution ism quite straightforward, we choose each empty cell, try values from 1 to 9, once validated, then we recursively call the solver function, until there is no other empty cell.

Code

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class Solution(object):
    def solveSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        self.numbers = ["1", "2", "3", "4", "5", "6", "7", "8", "9"]
        self.solveSu(board)
     
    def solveSu(self, board):
        for i in range(9):
            for j in range(9):
                if board[i][j] == '.':
                    for k in self.numbers:
                        if self.validateSudoku(board, k, i, j):
                            board[i][j] = k
                            if self.solveSu(board):
                                return True
                            else:
                                board[i][j] = '.'
                    return False
        
        return True
        
    def validateSudoku(self, board, k, i, j):
        for index in range(9):
            if board[i][index] == k:
                return False
            
        for index in range(9):
            if board[index][j] == k:
                return False
        
        start_row = int(i/3)*3
        start_col = int(j/3)*3
        for index_row in range(start_row, start_row+3):
            for index_col in range(start_col, start_col+3):
                if board[index_row][index_col] == k:
                    return False
        
        return True

Improvement

A little trick to make the backtracing faster is accelerate the validation process, by using hash table in each row, each column, each block.

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class Solution(object):
    def solveSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        self.numbers = ["1", "2", "3", "4", "5", "6", "7", "8", "9"]
        self.row_hash = [{} for j in range(9)]
        self.col_hash = [{} for j in range(9)]
        self.block_hash = [{} for j in range(9)]
        
        for i in range(9):
            for j in range(9):
                if board[i][j] != '.':
                    value = board[i][j]
                    self.row_hash[i][value] = True
                    self.col_hash[j][value] = True
                    self.block_hash[int(i/3)*3+(int(j/3))][value] = True
        self.solveSu(board)
        
     
    def solveSu(self, board):
        for i in range(9):
            for j in range(9):
                if board[i][j] == '.':
                    for k in self.numbers:
                        if self.validateSudoku(board, k, i, j):
                            board[i][j] = k
                            self.row_hash[i][k] = True
                            self.col_hash[j][k] = True
                            self.block_hash[int(i/3)*3+(int(j/3))][k] = True
                            if self.solveSu(board):
                                return True
                            else:
                                board[i][j] = '.'
                                self.row_hash[i][k] = False
                                self.col_hash[j][k] = False
                                self.block_hash[int(i/3)*3+(int(j/3))][k] = False
                    return False
        
        return True
        
    def validateSudoku(self, board, k, i, j):
        if self.row_hash[i].get(k) or self.col_hash[j].get(k) or self.block_hash[int(i/3)*3+(int(j/3))].get(k):
            return False
        
        
        return True

Experiment