Validity of a given Tic-Tac-Toe board configuration

A Tic-Tac-Toe board is given after some moves are played. Find out if the given board is valid, i.e., is it possible to reach this board position after some moves or not.
Note that every arbitrary filled grid of 9 spaces isn’t valid e.g. a grid filled with 3 X and 6 O isn’t valid situation because each player needs to take alternate turns.

Input is given as a 1D array of size 9. 

Examples:

Input: board[] =  {'X', 'X', 'O', 
                   'O', 'O', 'X',
                   'X', 'O', 'X'};
Output: Valid

Input: board[] =  {'O', 'X', 'X', 
                   'O', 'X', 'X',
                   'O', 'O', 'X'};
Output: Invalid
(Both X and O cannot win)

Input: board[] =  {'O', 'X', ' ', 
                   ' ', ' ', ' ',
                   ' ', ' ', ' '};
Output: Valid
(Valid board with only two moves played)

Basically, to find the validity of an input grid, we can think of the conditions when an input grid is invalid. Let no. of “X”s be countX and no. of “O”s be countO. Since we know that the game starts with X, a given grid of Tic-Tac-Toe game would be definitely invalid if following two conditions meet 

1. countX != countO AND 
2. countX != countO + 1
   • Since “X” is always the first move, second condition is also required.
   • Now does it mean that all the remaining board positions are valid one? The answer is NO. Think of the cases when input grid is such that both X and O are making straight lines. This is also not
   • valid position because the game ends when one player wins. So we need to check the following condition as well 
3. If input grid shows that both the players are in winning situation, it’s an invalid position. 
4. If input grid shows that the player with O has put a straight-line (i.e. is in win condition) and countX != countO, it’s an invalid position. The reason is that O plays his move only after X plays his
   • move. Since X has started the game, O would win when both X and O has played equal no. of moves. 
5. If input grid shows that X is in winning condition than xCount must be one greater that oCount.
   • Armed with above conditions i.e. a), b), c) and d), we can now easily formulate an algorithm/program to check the validity of a given Tic-Tac-Toe board position. 

1)  countX == countO or countX == countO + 1
2)  If O is in win condition then check 
     a)     If X also wins, not valid
     b)     If xbox != obox , not valid
3)  If X is in win condition then check if xCount is
     one more than oCount or not  

Another way to find the validity of a given board is using ‘inverse method’ i.e. rule out all the possibilities when a given board is invalid.

C++

// C++ program to check whether a given tic tac toe
// board is valid or not
#include <iostream>
using namespace std;

// This matrix is used to find indexes to check all
// possible winning triplets in board[0..8]
int win[8][3] = {{0, 1, 2}, // Check first row.
                {3, 4, 5}, // Check second Row
                {6, 7, 8}, // Check third Row
                {0, 3, 6}, // Check first column
                {1, 4, 7}, // Check second Column
                {2, 5, 8}, // Check third Column
                {0, 4, 8}, // Check first Diagonal
                {2, 4, 6}}; // Check second Diagonal

// Returns true if character 'c' wins. c can be either
// 'X' or 'O'
bool isCWin(char *board, char c)
{
    // Check all possible winning combinations
    for (int i=0; i<8; i++)
        if (board[win[i][0]] == c &&
            board[win[i][1]] == c &&
            board[win[i][2]] == c )
            return true;
    return false;
}

// Returns true if given board is valid, else returns false
bool isValid(char board[9])
{
    // Count number of 'X' and 'O' in the given board
    int xCount=0, oCount=0;
    for (int i=0; i<9; i++)
    {
    if (board[i]=='X') xCount++;
    if (board[i]=='O') oCount++;
    }

    // Board can be valid only if either xCount and oCount
    // is same or count is one more than oCount
    if (xCount==oCount || xCount==oCount+1)
    {
        // Check if 'O' is winner
        if (isCWin(board, 'O'))
        {
            // Check if 'X' is also winner, then
            // return false
            if (isCWin(board, 'X'))
                return false;

            // Else return true xCount and yCount are same
            return (xCount == oCount);
        }

        // If 'X' wins, then count of X must be greater
        if (isCWin(board, 'X') && xCount != oCount + 1)
        return false;

        // If 'O' is not winner, then return true
        return true;
    }
    return false;
}

// Driver program
int main()
{
char board[] = {'X', 'X', 'O',
                'O', 'O', 'X',
                'X', 'O', 'X'};
(isValid(board))? cout << "Given board is valid":
                    cout << "Given board is not valid";
return 0;
}

Java

// Java program to check whether a given tic tac toe
// board is valid or not
import java.io.*;
class GFG {

// This matrix is used to find indexes to check all
// possible winning triplets in board[0..8]
    static int win[][] = {{0, 1, 2}, // Check first row.
    {3, 4, 5}, // Check second Row
    {6, 7, 8}, // Check third Row
    {0, 3, 6}, // Check first column
    {1, 4, 7}, // Check second Column
    {2, 5, 8}, // Check third Column
    {0, 4, 8}, // Check first Diagonal
    {2, 4, 6}}; // Check second Diagonal

// Returns true if character 'c' wins. c can be either
// 'X' or 'O'
    static boolean isCWin(char[] board, char c) {
        // Check all possible winning combinations
        for (int i = 0; i < 8; i++) {
            if (board[win[i][0]] == c
                    && board[win[i][1]] == c
                    && board[win[i][2]] == c) {
                return true;
            }
        }
        return false;
    }

// Returns true if given board is valid, else returns false
    static boolean isValid(char board[]) {
        // Count number of 'X' and 'O' in the given board
        int xCount = 0, oCount = 0;
        for (int i = 0; i < 9; i++) {
            if (board[i] == 'X') {
                xCount++;
            }
            if (board[i] == 'O') {
                oCount++;
            }
        }

        // Board can be valid only if either xCount and oCount
        // is same or count is one more than oCount
        if (xCount == oCount || xCount == oCount + 1) {
            // Check if 'O' is winner
            if (isCWin(board, 'O')) {
                // Check if 'X' is also winner, then
                // return false
                if (isCWin(board, 'X')) {
                    return false;
                }

                // Else return true xCount and yCount are same
                return (xCount == oCount);
            }

            // If 'X' wins, then count of X must be greater
            if (isCWin(board, 'X') && xCount != oCount + 1) {
                return false;
            }

            // If 'O' is not winner, then return true
            return true;
        }
        return false;
    }

// Driver program
    public static void main(String[] args) {
        char board[] = {'X', 'X', 'O', 'O', 'O', 'X', 'X', 'O', 'X'};

        if ((isValid(board))) {
            System.out.println("Given board is valid");
        } else {
            System.out.println("Given board is not valid");
        }
    }
}
//this code contributed by PrinciRaj1992

Python3

# Python3 program to check whether a given tic tac toe
# board is valid or not

# Returns true if char wins. Char can be either
# 'X' or 'O'
def win_check(arr, char):
    # Check all possible winning combinations
    matches = [[0, 1, 2], [3, 4, 5],
               [6, 7, 8], [0, 3, 6],
               [1, 4, 7], [2, 5, 8],
               [0, 4, 8], [2, 4, 6]]

    for i in range(8):
        if(arr[(matches[i][0])] == char and
            arr[(matches[i][1])] == char and
            arr[(matches[i][2])] == char):
            return True
    return False

def is_valid(arr):
    # Count number of 'X' and 'O' in the given board
    xcount = arr.count('X')
    ocount = arr.count('O')

    # Board can be valid only if either xcount and ocount
    # is same or count is one more than oCount
    if(xcount == ocount+1 or xcount == ocount):
        # Check if O wins
        if win_check(arr, 'O'):
            # Check if X wins, At a given point only one can win,
            # if X also wins then return Invalid
            if win_check(arr, 'X'):
                return "Invalid"

            # O can only win if xcount == ocount in case where whole
            # board has values in each position.
            if xcount == ocount:
                return "Valid"

        # If X wins then it should be xc == oc + 1,
        # If not return Invalid
        if win_check(arr, 'X') and xcount != ocount+1:
            return "Invalid"

        # if O is not the winner return Valid
        if not win_check(arr, 'O'):
            return "valid"

    # If nothing above matches return invalid
    return "Invalid"


# Driver Code
arr = ['X', 'X', 'O',
       'O', 'O', 'X',
       'X', 'O', 'X']
print("Given board is " + is_valid(arr))

C#

// C# program to check whether a given
// tic tac toe board is valid or not
using System;

class GFG
{

// This matrix is used to find indexes
// to check all possible winning triplets
// in board[0..8]
public static int[][] win = new int[][]
{
    new int[] {0, 1, 2},
    new int[] {3, 4, 5},
    new int[] {6, 7, 8},
    new int[] {0, 3, 6},
    new int[] {1, 4, 7},
    new int[] {2, 5, 8},
    new int[] {0, 4, 8},
    new int[] {2, 4, 6}
};

// Returns true if character 'c'
// wins. c can be either 'X' or 'O'
public static bool isCWin(char[] board,
                          char c)
{
    // Check all possible winning
    // combinations
    for (int i = 0; i < 8; i++)
    {
        if (board[win[i][0]] == c &&
            board[win[i][1]] == c &&
            board[win[i][2]] == c)
        {
            return true;
        }
    }
    return false;
}

// Returns true if given board
// is valid, else returns false
public static bool isValid(char[] board)
{
    // Count number of 'X' and
    // 'O' in the given board
    int xCount = 0, oCount = 0;
    for (int i = 0; i < 9; i++)
    {
        if (board[i] == 'X')
        {
            xCount++;
        }
        if (board[i] == 'O')
        {
            oCount++;
        }
    }

    // Board can be valid only if either
    // xCount and oCount is same or count
    // is one more than oCount
    if (xCount == oCount ||
        xCount == oCount + 1)
    {
        // Check if 'O' is winner
        if (isCWin(board, 'O'))
        {
            // Check if 'X' is also winner,
            // then return false
            if (isCWin(board, 'X'))
            {
                return false;
            }

            // Else return true xCount
            // and yCount are same
            return (xCount == oCount);
        }

        // If 'X' wins, then count of
        // X must be greater
        if (isCWin(board, 'X') &&
            xCount != oCount + 1)
        {
            return false;
        }

        // If 'O' is not winner,
        // then return true
        return true;
    }
    return false;
}

// Driver Code
public static void Main(string[] args)
{
    char[] board = new char[] {'X', 'X', 'O', 'O', 'O',
                                    'X', 'X', 'O', 'X'};

    if ((isValid(board)))
    {
        Console.WriteLine("Given board is valid");
    }
    else
    {
        Console.WriteLine("Given board is not valid");
    }
}
}

// This code is contributed by Shrikant13

PHP

<?php
// PHP program to check whether a given
// tic tac toe board is valid or not

// This matrix is used to find indexes
// to check all possible winning triplets
// in board[0..8]

// Returns true if character 'c' wins.
// c can be either 'X' or 'O'
function isCWin($board, $c)
{
    $win = array(array(0, 1, 2), // Check first row.
                 array(3, 4, 5), // Check second Row
                 array(6, 7, 8), // Check third Row
                 array(0, 3, 6), // Check first column
                 array(1, 4, 7), // Check second Column
                 array(2, 5, 8), // Check third Column
                 array(0, 4, 8), // Check first Diagonal
                 array(2, 4, 6)); // Check second Diagonal

    // Check all possible winning combinations
    for ($i = 0; $i < 8; $i++)
        if ($board[$win[$i][0]] == $c &&
            $board[$win[$i][1]] == $c &&
            $board[$win[$i][2]] == $c )
            return true;
    return false;
}

// Returns true if given board is
// valid, else returns false
function isValid(&$board)
{
    // Count number of 'X' and 'O'
    // in the given board
    $xCount = 0;
    $oCount = 0;
    for ($i = 0; $i < 9; $i++)
    {
        if ($board[$i] == 'X') $xCount++;
        if ($board[$i] == 'O') $oCount++;
    }

    // Board can be valid only if either
    // xCount and oCount is same or count
    // is one more than oCount
    if ($xCount == $oCount || $xCount == $oCount + 1)
    {
        // Check if 'O' is winner
        if (isCWin($board, 'O'))
        {
            // Check if 'X' is also winner,
            // then return false
            if (isCWin($board, 'X'))
                return false;

            // Else return true xCount and
            // yCount are same
            return ($xCount == $oCount);
        }

        // If 'X' wins, then count of X
        // must be greater
        if (isCWin($board, 'X') &&
                   $xCount != $oCount + 1)
        return false;

        // If 'O' is not winner, then
        // return true
        return true;
    }
    return false;
}

// Driver Code
$board = array('X', 'X', 'O','O',
               'O', 'X','X', 'O', 'X');
if(isValid($board))
    echo("Given board is valid");
else
    echo ("Given board is not valid");

// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>
// Javascript program to check whether a given
// tic tac toe board is valid or not

// This matrix is used to find indexes
// to check all possible winning triplets
// in board[0..8]

// Returns true if character 'c' wins.
// c can be either 'X' or 'O'
function isCWin(board, c)
{
    let win = new Array(new Array(0, 1, 2), // Check first row.
                new Array(3, 4, 5), // Check second Row
                new Array(6, 7, 8), // Check third Row
                new Array(0, 3, 6), // Check first column
                new Array(1, 4, 7), // Check second Column
                new Array(2, 5, 8), // Check third Column
                new Array(0, 4, 8), // Check first Diagonal
                new Array(2, 4, 6)); // Check second Diagonal

    // Check all possible winning combinations
    for (let i = 0; i < 8; i++)
        if (board[win[i][0]] == c &&
            board[win[i][1]] == c &&
            board[win[i][2]] == c )
            return true;
    return false;
}

// Returns true if given board is
// valid, else returns false
function isValid(board)
{
    // Count number of 'X' and 'O'
    // in the given board
    let xCount = 0;
    let oCount = 0;
    for (let i = 0; i < 9; i++)
    {
        if (board[i] == 'X') xCount++;
        if (board[i] == 'O') oCount++;
    }

    // Board can be valid only if either
    // xCount and oCount is same or count
    // is one more than oCount
    if (xCount == oCount || xCount == oCount + 1)
    {
        // Check if 'O' is winner
        if (isCWin(board, 'O'))
        {
            // Check if 'X' is also winner,
            // then return false
            if (isCWin(board, 'X'))
                return false;

            // Else return true xCount and
            // yCount are same
            return (xCount == oCount);
        }

        // If 'X' wins, then count of X
        // must be greater
        if (isCWin(board, 'X') &&
                xCount != oCount + 1)
        return false;

        // If 'O' is not winner, then
        // return true
        return true;
    }
    return false;
}

// Driver Code
let board = new Array('X', 'X', 'O','O',
            'O', 'X','X', 'O', 'X');
if(isValid(board))
    document.write("Given board is valid");
else
    document.write("Given board is not valid");

// This code is contributed
// by Saurabh Jaiswal
</script>

Output

Given board is valid

Time complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.

Thanks to Utkarsh for suggesting this solution. This article is contributed by Aarti_Rathi and Utkarsh. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above