Check if given Sudoku board configuration is valid or not

Given a Sudoku Board configuration, check whether it is valid or not.

Examples:

Input: 
    [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

The basic idea is to check whether each row, column, and the 3×3 box is valid or not on the basis of following points:



  • The Sudoku board could be partially filled, where empty cells are filled with the character ‘.’.
  • An empty Sudoku board is also valid.
  • A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

Below is the implementation of above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// Program to check whether given sudoku 
// board is valid or not
#include <bits/stdc++.h>
using namespace std;
  
// Checks whether there is any duplicate 
// in current row or not
bool notInRow(char arr[][9], int row)
{
    // Set to store characters seen so far.
    set<char> st;
  
    for (int i = 0; i < 9; i++) {
  
        // If already encountered before, return false
        if (st.find(arr[row][i]) != st.end())
            return false;
  
        // If it is not an empty cell, insert value
        // at the current cell in the set
        if (arr[row][i] != '.')
            st.insert(arr[row][i]);
    }
    return true;
}
  
// Checks whether there is any duplicate
// in current column or not.
bool notInCol(char arr[][9], int col)
{
    set<char> st;
  
    for (int i = 0; i < 9; i++) {
  
        // If already encountered before, return false
        if (st.find(arr[i][col]) != st.end())
            return false;
  
        // If it is not an empty cell,
        // insert value at the current cell in the set
        if (arr[i][col] != '.')
            st.insert(arr[i][col]);
    }
    return true;
}
  
// Checks whether there is any duplicate
// in current 3x3 box or not.
bool notInBox(char arr[][9], int startRow, int startCol)
{
    set<char> st;
  
    for (int row = 0; row < 3; row++) {
        for (int col = 0; col < 3; col++) {
            char curr = arr[row + startRow][col + startCol];
  
            // If already encountered before, return false
            if (st.find(curr) != st.end())
                return false;
  
            // If it is not an empty cell,
            // insert value at current cell in set
            if (curr != '.')
                st.insert(curr);
        }
    }
    return true;
}
  
// Checks whether current row and current column and
// current 3x3 box is valid or not
bool isValid(char arr[][9], int row, int col)
{
    return notInRow(arr, row) && notInCol(arr, col) &&
           notInBox(arr, row - row % 3, col - col % 3);
}
  
bool isValidConfig(char arr[][9], int n)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
  
            // If current row or current column or
            // current 3x3 box is not valid, return false
            if (!isValid(arr, i, j))
                return false;
        }
    }
    return true;
}
  
// Drivers code
int main()
{
    char board[9][9] = { { '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' } };
  
    cout << (isValidConfig(board, 9) ? "YES\n" : "NO\n");
    return 0;
}

chevron_right






                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 program to check whether  


# given sudoku board is valid or not  


  


# Checks whether there is any  


# duplicate in current row or not  


def notInRow(arr, row):  


  


    # Set to store characters seen so far.  


    st = set()  


  


    for i in range(0, 9):  


  


        # If already encountered before,  


        # return false  


        if arr[row][i] in st:  


            return False


  


        # If it is not an empty cell, insert value  


        # at the current cell in the set  


        if arr[row][i] != '.':  


            st.add(arr[row][i])  


      


    return True


  


# Checks whether there is any  


# duplicate in current column or not.  


def notInCol(arr, col):  


  


    st = set()  


  


    for i in range(0, 9):  


  


        # If already encountered before, 


        # return false  


        if arr[i][col] in st: 


            return False


  


        # If it is not an empty cell, insert  


        # value at the current cell in the set  


        if arr[i][col] != '.':  


            st.add(arr[i][col])  


      


    return True


  


# Checks whether there is any duplicate  


# in current 3x3 box or not.  


def notInBox(arr, startRow, startCol):  


  


    st = set()  


  


    for row in range(0, 3):  


        for col in range(0, 3):  


            curr = arr[row + startRow][col + startCol]  


  


            # If already encountered before,  


            # return false  


            if curr in st:  


                return False


  


            # If it is not an empty cell,  


            # insert value at current cell in set  


            if curr != '.':  


                st.add(curr)  


          


    return True


  


# Checks whether current row and current  


# column and current 3x3 box is valid or not  


def isValid(arr, row, col):  


  


    return (notInRow(arr, row) and notInCol(arr, col) and


            notInBox(arr, row - row % 3, col - col % 3))  


  


def isValidConfig(arr, n):  


  


    for i in range(0, n):  


        for j in range(0, n):  


  


            # If current row or current column or  


            # current 3x3 box is not valid, return false  


            if not isValid(arr, i, j):  


                return False


          


    return True


  


# Drivers code  


if __name__ == "__main__": 


  


    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' ]] 


          


    if isValidConfig(board, 9):  


        print("YES") 


    else: 


        print("NO") 


  


# This code is contributed by Rituraj Jain 



















                        chevron_right







                        filter_none











Output:


YES






          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
Aashish Chauhan
A Coding Enthusiast
Rails Developer
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.


Improved By :  rituraj_jain


					
		

    Article Tags : 
Matrix

           	
Practice Tags : 
Matrix
STL

        

         

         
 
                2

                

        



        
        
        
        
        

    

            

            3
            

            

             
                 Based on 7 vote(s)
                    

            
            
            
            
            
        

        
        

        

        

        

        

        

        


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.