# 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.

```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)
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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
a) countX != countO AND
b) 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
c) If input grid shows that both the players are in winning situation, it’s an invalid position.
d) 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.
e) 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++ 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 wining 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 xount 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;
}
```

Output:

`Given board is valid`

Thanks to Utkarsh for suggesting this solution here.

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