Find Winner on a Tic Tac Toe Game

Problem

Tic-tac-toe is played by two players A and B on a 3 x 3 grid. The rules of Tic-Tac-Toe are:

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 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.

Given a 2D integer array moves where moves[i] = [rowi, coli] indicates that the ith move will be played on grid[rowi][coli]. 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 (i.e., 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, they always play first.

Example 2:

Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: B wins.

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.

Constraints:

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

Solution (Java)

class Solution {
    public String tictactoe(int[][] moves) {
        String[][] board = new String[3][3];
        for (int i = 0; i < moves.length; i++) {
            if (i % 2 == 0) {
                board[moves[i][0]][moves[i][1]] = "X";
            } else {
                board[moves[i][0]][moves[i][1]] = "O";
            }
            if (i > 3 && !wins(board).equals("")) {
                return wins(board);
            }
        }
        return moves.length == 9 ? "Draw" : "Pending";
    }

    private String wins(String[][] board) {
        for (int i = 0; i < 3; i++) {
            if (board[i][0] == null) {
                break;
            }
            String str = board[i][0];
            if (str.equals(board[i][1]) && str.equals(board[i][2])) {
                return getWinner(str);
            }
        }

        for (int j = 0; j < 3; j++) {
            if (board[0][j] == null) {
                break;
            }
            String str = board[0][j];
            if (str.equals(board[1][j]) && str.equals(board[2][j])) {
                return getWinner(str);
            }
        }

        if (board[1][1] == null) {
            return "";
        }
        String str = board[1][1];
        if (str.equals(board[0][0]) && str.equals(board[2][2])
                || (str.equals(board[0][2]) && str.equals(board[2][0]))) {
            return getWinner(str);
        }
        return "";
    }

    private String getWinner(String str) {
        if (str.equals("X")) {
            return "A";
        } else {
            return "B";
        }
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).