# Valid Tic-Tac-Toe State Problem

## Description

LeetCode Problem 794.

Given a Tic-Tac-Toe board as a string array board, return true if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.

The board is a 3 x 3 array that consists of characters ‘ ‘, ‘X’, and ‘O’. The ‘ ‘ character represents an empty square.

Here are the rules of Tic-Tac-Toe:

• Players take turns placing characters into empty squares ‘ ‘.
• The first player always places ‘X’ characters, while the second player always places ‘O’ characters.
• ‘X’ and ‘O’ characters are always placed into empty squares, never filled ones.
• The game ends when there are three of the same (non-empty) character filling any row, column, or diagonal.
• The game also ends if all squares are non-empty.
• No more moves can be played if the game is over.

Example 1:

``````1
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Input: board = ["O  ","   ","   "]
Output: false
Explanation: The first player always plays "X".
``````

Example 2:

``````1
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Input: board = ["XOX"," X ","   "]
Output: false
Explanation: Players take turns making moves.
``````

Example 3:

``````1
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Input: board = ["XXX","   ","OOO"]
Output: false
``````

Example 4:

``````1
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Input: board = ["XOX","O O","XOX"]
Output: true
``````

Constraints:

• board.length == 3
• board[i].length == 3
• board[i][j] is either ‘X’, ‘O’, or ‘ ‘.

## Sample C++ Code

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class Solution {
public:
bool iswin(vector<string> &board, char c) {
return ((board[0][0] == c && board[0][1] == c && board[0][2] == c) ||
(board[1][0] == c && board[1][1] == c && board[1][2] == c) ||
(board[2][0] == c && board[2][1] == c && board[2][2] == c) ||
(board[0][0] == c && board[1][0] == c && board[2][0] == c) ||
(board[0][1] == c && board[1][1] == c && board[2][1] == c) ||
(board[0][2] == c && board[1][2] == c && board[2][2] == c) ||
(board[0][0] == c && board[1][1] == c && board[2][2] == c) ||
(board[0][2] == c && board[1][1] == c && board[2][0] == c));
}

bool validTicTacToe(vector<string>& board) {
int xcnt = 0, ocnt = 0;
for (string s: board) {
for (char c: s) {
if (c == 'X')
xcnt ++;
else if (c == 'O')
ocnt ++;
}
}
if (xcnt != ocnt && xcnt != ocnt + 1)
// X can have equal or only 1 more move than O
return false;
if (iswin(board, 'X') && xcnt != ocnt + 1)
// X will win by having one additional move
return false;
if (iswin(board, 'O') && xcnt != ocnt)
// since X played first, O can win only when both have made same number of moves
return false;
return true;
}
};
``````