Tic-Tac-Toe

Last updated: 8th Sept, 2020

Problem Statement

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.
Note : The game starts with X

Example 1:

2
X X O O X O O O O
X X O O O X X O X

>> Invalid
>> Valid

Example 2:

1
O X X X O X O O X

>> Invalid
>> Valid

Approach:

In this problem we check the validity of the given inputs of a tic-tac-toe game. The game involves two players 'X' and 'O' who take turns marking the spaces in a 3×3 grid. There are 8 strategies in order to win this game.
Consider player 'X', if 'X' occupies one of the row or one of the column or occupies the diagonals of grid then 'X' wins. If 'X' wins then the game ends which also applies to 'O'.

CONDITIONS TO CHECK IF INPUTS ARE INVALID:

If both player 'X' and 'O' wins
When 'O' takes a chance after 'X' wins or vice versa
When number of 'X' exceed number of 'O' by two or more
When number of 'O's are greater than 'X'

CHECKING THE VALIDITY:

Number of 'X' exceeds number of 'O' if 'X' wins.
Number of 'X' and 'O' equal only if 'O' wins.
Number of 'X' exceed number of 'O' by 1 not greater than that.

POSSIBILITIES OF WINNING A GAME:

'X' occupies one of the row or 'O' occupies one of the row.
'X' occupies one of the column or 'O' occupies one of the column.
'X' occupies one of the diagonals in the grid or 'O' occupies one of the diagonals in the grid.

Complexity Analysis :

Time Complexity- O(n^2)
Space Complexity- O(n)

Pseudocode :
PROGRAM ticTacToe
// Possibilities() is a function to represent possible ways of winnning
      INPUT: Read T, a
       SET countX = 0
       SET countY = 0
       FOR each element in a
             IF element = 0 THEN
                   INCREMENT countX

             ELSE
                   INCREMENT countO
             END IF
       IF countO > countX or countX > countO + 1 THEN
             PRINT Invalid
       ELSE IF Possibilities(a, 'X') and Possibilities(a, 'O') THEN
             PRINT Invalid
       ELSE IF Possibilities(a, 'O') THEN
             IF countX == countO THEN
                   PRINT Valid
             ELSE
                  PRINT Invalid.
      ELSE IF countO + 1 == countX or self.Possibilities(a, 'X') THEN
            PRINT Valid
      ELSE
            PRINT Invalid
      END IF
       END FOR
END

Python Code

class ticTacToe:

    def Possibilities(self, a, b):

        return ((a[0] == b and a[1] == b and a[2] == a[0]) or
            (a[3] == b and a[4] == b and a[5] == b) or
            (a[6] == b and a[7] == b and a[8] == b) or
            (a[0] == b and a[3] == b and a[6] == b) or
            (a[1] == b and a[4] == b and a[7] == b) or
            (a[2] == b and a[5] == b and a[8] == b) or
            (a[0] == b and a[4] == b and a[8] == b) or
            (a[2] == b and a[4] == b and a[6] == b))

    def validity(self, T):
        if 1 <= T <= 100:
            for i in range(T):
                countX = 0
                countO = 0
                a = input().split()
                for t in a:
                    if t == 'X':
                        countX += 1
                    else:
                        countO += 1
                if countO > (countX) or (countX > countO + 1) :
                    print("Invalid")
                elif self.Possibilities(a, 'X') and self.Possibilities(a, 'O'):
                    print("Invalid")
                elif self.Possibilities(a, 'O'):
                    if countX == countO:
                        print('Valid')
                    else:
                        print("Invalid")
                elif countO + 1 == countX or self.Possibilities(a, 'X'):
                    print("Valid")
                else:
                    print("Invalid")

chec = ticTacToe()
chec.validity(int(input()))