Formatted question description: https://leetcode.ca/all/1275.html

1275. Find Winner on a Tic Tac Toe Game (Easy)

Tic-tac-toe is played by two players A and B on a 3 x 3 grid.

Here are the rules of Tic-Tac-Toe:

• Players take turns placing characters into empty squares (" ").
• The first player A always places "X" characters, while the second player B always places "O" characters.
• "X" and "O" characters are always placed into empty squares, never on 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.

Given an array moves where each element is another array of size 2 corresponding to the row and column of the grid where they mark their respective character in the order in which A and B play.

Return the winner of the game if it exists (A or B), in case the game ends in a draw return "Draw", if there are still movements to play return "Pending".

You can assume that moves is valid (It follows the rules of Tic-Tac-Toe), the grid is initially empty and A will play first.

 

Example 1:

Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
Output: "A"
Explanation: "A" wins, he always plays first.
"X  "    "X  "    "X  "    "X  "    "X  "
"   " -> "   " -> " X " -> " X " -> " X "
"   "    "O  "    "O  "    "OO "    "OOX"

Example 2:

Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: "B" wins.
"X  "    "X  "    "XX "    "XXO"    "XXO"    "XXO"
"   " -> " O " -> " O " -> " O " -> "XO " -> "XO " 
"   "    "   "    "   "    "   "    "   "    "O  "

Example 3:

Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
Output: "Draw"
Explanation: The game ends in a draw since there are no moves to make.
"XXO"
"OOX"
"XOX"

Example 4:

Input: moves = [[0,0],[1,1]]
Output: "Pending"
Explanation: The game has not finished yet.
"X  "
" O "
"   "

 

Constraints:

• 1 <= moves.length <= 9
• moves[i].length == 2
• 0 <= moves[i][j] <= 2
• There are no repeated elements on moves.
• moves follow the rules of tic tac toe.

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/

// Time: O(1)
// Space: O(1)
class Solution {
    string check(int A[3][3]) {
        for (int i = 0; i < 3; ++i) {
            int j = 1;
            while (j < 3 && A[i][0] != 0 && A[i][j] == A[i][0]) ++j;
            if (j == 3) return A[i][0] == 1 ? "A" : "B";
            j = 1;
            while (j < 3 && A[0][i] != 0 && A[j][i] == A[0][i]) ++j; 
            if (j == 3) return A[0][i] == 1 ? "A" : "B";
        }
        int i = 1;
        while (i < 3 && A[0][0] != 0 && A[i][i] == A[0][0]) ++i; 
        if (i == 3) return A[0][0] == 1 ? "A" : "B";
        i = 1;
        while (i < 3 && A[0][2] != 0 && A[i][2 - i] == A[0][2]) ++i;
        if (i == 3) return A[0][2] == 1 ? "A" : "B";
        for (int i = 0; i < 3; ++i) {
            for (int j = 0; j < 3; ++j) {
                if (A[i][j] == 0) return "Pending";
            }
        }
        return "Draw";
    }
public:
    string tictactoe(vector<vector<int>>& moves) {
        int A[3][3] = {}, ch = 1;
        for (auto &m : moves) {
            A[m[0]][m[1]] = ch;
            ch = ch == 1 ? 2 : 1;
        }
        return check(A);
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/

// Time: O(1)
// Space: O(1)
class Solution {
public:
    string tictactoe(vector<vector<int>>& moves) {
        int A[8] = {}, B[8] = {};
        for (int i = 0; i < moves.size(); ++i) {
            int r = moves[i][0], c = moves[i][1];
            if (i % 2 == 0 && (++A[r] == 3 || ++A[3 + c] == 3 || (r == c && ++A[6] == 3) || (r == 2 - c && ++A[7] == 3))) return "A";
            if (i % 2 && (++B[r] == 3 || ++B[3 + c] == 3 || (r == c && ++B[6] == 3) || (r == 2 - c && ++B[7] == 3))) return "B";
        }
        return moves.size() == 9 ? "Draw" : "Pending";
    }
};

Java

class Solution {
    public String tictactoe(int[][] moves) {
        final char[] SIGNALS = {'X', 'O'};
        final int SIDE = 3;
        char[][] board = new char[SIDE][SIDE];
        int[][] countRows = new int[2][SIDE];
        int[][] countColumns = new int[2][SIDE];
        int[][] countDiagonals = new int[2][2];
        int length = moves.length;
        for (int i = 0; i < length; i++) {
            int player = i % 2;
            int[] move = moves[i];
            int row = move[0], column = move[1];
            board[row][column] = SIGNALS[player];
            int countRow = countRows[player][row];
            countRow++;
            countRows[player][row] = countRow;
            if (countRow == SIDE)
                return player == 0 ? "A" : "B";
            int countColumn = countColumns[player][column];
            countColumn++;
            countColumns[player][column] = countColumn;
            if (countColumn == SIDE)
                return player == 0 ? "A" : "B";
            if (row == column) {
                int countDiagonal = countDiagonals[player][0];
                countDiagonal++;
                countDiagonals[player][0] = countDiagonal;
                if (countDiagonal == SIDE)
                    return player == 0 ? "A" : "B";
            }
            if (row + column == SIDE - 1) {
                int countDiagonal = countDiagonals[player][1];
                countDiagonal++;
                countDiagonals[player][1] = countDiagonal;
                if (countDiagonal == SIDE)
                    return player == 0 ? "A" : "B";
            }
        }
        if (length == 9)
            return "Draw";
        else
            return "Pending";
    }
}

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