Formatted question description: https://leetcode.ca/all/794.html

# 794. Valid Tic-Tac-Toe State (Medium)

A Tic-Tac-Toe board is given 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, and 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 3 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:
Input: board = [“O  “, “   ”, “   ”]
Output: false
Explanation: The first player always plays “X”.

Example 2:
Input: board = [“XOX”, “ X “, “ “]
Output: false
Explanation: Players take turns making moves.

Example 3:
Input: board = [“XXX”, “ “, “OOO”]
Output: false

Example 4:
Input: board = [“XOX”, “O O”, “XOX”]
Output: true

Note:

• board is a length-3 array of strings, where each string board[i] has length 3.
• Each board[i][j] is a character in the set {" ", "X", "O"}.

## Solution 1.

// OJ: https://leetcode.com/problems/valid-tic-tac-toe-state/
// Time: O(1)
// Space: O(1)
class Solution {
private:
int count(vector<string>& board, char target){
int cnt = 0;
for (auto row : board) {
for (char c : row) {
if (c == target) ++cnt;
}
}
return cnt;
}
bool testWin(vector<string>& board, char target) {
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (board[i][j] != target) break;
if (j == 2) return true;
}
}
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (board[j][i] != target) break;
if (j == 2) return true;
}
}
for (int i = 0; i < 3; ++i) {
if (board[i][i] != target) break;
if (i == 2) return true;
}
for (int i = 0; i < 3; ++i) {
if (board[i][2 - i] != target) break;
if (i == 2) return true;
}
return false;
}
public:
bool validTicTacToe(vector<string>& board) {
int xCnt = count(board, 'X');
int oCnt = count(board, 'O');
if (xCnt != oCnt && xCnt != oCnt + 1) return false;
bool xWin = testWin(board, 'X');
bool oWin = testWin(board, 'O');
if (xWin && oWin) return false;
if (!xWin && !oWin) return true;
return (xWin && xCnt == oCnt + 1) || (oWin && xCnt == oCnt);
}
};


Java

• class Solution {
public boolean validTicTacToe(String[] board) {
int xCount = 0, oCount = 0;
for (String row: board) {
for (char c: row.toCharArray()) {
if (c == 'X')
xCount++;
if (c == 'O')
oCount++;
}
}
if (oCount != xCount && oCount != xCount - 1)
return false;
if (win(board, 'X') && oCount != xCount - 1)
return false;
if (win(board, 'O') && oCount != xCount)
return false;
return true;
}

public boolean win(String[] B, char P) {
for (int i = 0; i < 3; i++) {
if (P == B.charAt(i) && P == B.charAt(i) && P == B.charAt(i))
return true;
if (P == B[i].charAt(0) && P == B[i].charAt(1) && P == B[i].charAt(2))
return true;
}
if (P == B.charAt(0) && P == B.charAt(1) && P == B.charAt(2))
return true;
if (P == B.charAt(2) && P == B.charAt(1) && P == B.charAt(0))
return true;
return false;
}
}

• // OJ: https://leetcode.com/problems/valid-tic-tac-toe-state/
// Time: O(1)
// Space: O(1)
class Solution {
private:
int count(vector<string>& board, char target){
int cnt = 0;
for (auto row : board) {
for (char c : row) {
if (c == target) ++cnt;
}
}
return cnt;
}
bool testWin(vector<string>& board, char target) {
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (board[i][j] != target) break;
if (j == 2) return true;
}
}
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (board[j][i] != target) break;
if (j == 2) return true;
}
}
for (int i = 0; i < 3; ++i) {
if (board[i][i] != target) break;
if (i == 2) return true;
}
for (int i = 0; i < 3; ++i) {
if (board[i][2 - i] != target) break;
if (i == 2) return true;
}
return false;
}
public:
bool validTicTacToe(vector<string>& board) {
int xCnt = count(board, 'X');
int oCnt = count(board, 'O');
if (xCnt != oCnt && xCnt != oCnt + 1) return false;
bool xWin = testWin(board, 'X');
bool oWin = testWin(board, 'O');
if (xWin && oWin) return false;
if (!xWin && !oWin) return true;
return (xWin && xCnt == oCnt + 1) || (oWin && xCnt == oCnt);
}
};

• class Solution:
def validTicTacToe(self, board):
"""
:type board: List[str]
:rtype: bool
"""
xCount, oCount = 0, 0
for i in range(3):
for j in range(3):
if board[i][j] == 'O':
oCount += 1
elif board[i][j] == 'X':
xCount += 1
if oCount != xCount and oCount != xCount - 1: return False
if oCount != xCount and self.win(board, 'O'): return False
if oCount != xCount - 1 and self.win(board, 'X'): return False
return True

def win(self, board, P):
# board is list[str]
# P is 'X' or 'O' for two players
for j in range(3):
if all(board[i][j] == P for i in range(3)): return True
if all(board[j][i] == P for i in range(3)): return True
if board == board == board == P: return True
if board == board == board == P: return True
return False