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[0].charAt(i) && P == B[1].charAt(i) && P == B[2].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[0].charAt(0) && P == B[1].charAt(1) && P == B[2].charAt(2))
            return true;
        if (P == B[0].charAt(2) && P == B[1].charAt(1) && P == B[2].charAt(0))
            return true;
        return false;
    }
}

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