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

https://leetcode.com/problems/valid-tic-tac-toe-state/

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"}`.

## Solutions

``````class Solution {

// This is a nasty problem, too many special cases, skip it.

public boolean validTicTacToe(String[] board) {
int diff = count(board, 'X') - count(board, 'O');

// Initially, count of X must be equal to or 1 more than O, otherwise not a
// valid Tic-Tac-Toe board
if (diff > 1 || diff < 0) {
return false;
}

if (hasWon(board, 'X')) {
return diff == 1;
}

if (hasWon(board, 'O')) {
return diff == 0;
}

return true;
}

private boolean hasWon(String[] board, char player) {
int size = board.length;

// search the winner horizontally
for (int i = 0; i < size; i++) {
int count = 0;
for (int j = 0; j < size && board[i].charAt(j) == player; j++) {
count++;
}

if (count == size) {
return true;
}
}

// search the winner vertically
for (int j = 0; j < size; j++) {
int count = 0;
for (int i = 0; i < size && board[i].charAt(j) == player; i++) {
count++;
}

if (count == size) {
return true;
}
}

// search the winner diagonally, from left top to right bottom corner
int count = 0;
for (int i = 0; i < size && board[i].charAt(i) == player; i++) {
count++;
}

if (count == size) {
return true;
}

// search the winner diagonally, from left bottom to right top corner
for (int i = 0; i < size && board[size-i-1].charAt(i) == player; i++) {
count++;
}

if (count == size) {
return true;
}

return false;
}

private int count(String[] board, char player) {
int count = 0;

for (String row : board) {
for (char c : row.toCharArray()) {
if (c != player) {
continue;
}

count++;
}
}

return count;
}
}
``````