Find Winner on a Tic Tac Toe Game

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.

Solution:

class Solution {
    public String tictactoe(int[][] moves) {
        int[][] board = new int[3][3];
        boolean x = true;
        for (int[] move : moves) {
            fill(board, move, x);
            x = !x;
        }
        String w = winner(board);
        if (w.equals("Pending") && moves.length == 9) return "Draw";
        return w;
    }
    
    private void fill(int[][] board, int[] move, boolean x) {
        int i = move[0], j = move[1];
        if (x) {
            board[i][j] = 1;
        } else {
            board[i][j] = 2;
        }
    }
    
    private String winner(int[][] board) {
        for (int i = 0; i < 3; i ++) {
            if (board[i][0] != 0 && board[i][0] == board[i][1] && board[i][1] == board[i][2]) {
                return board[i][0] == 1 ? "A" : "B";
            }
            if (board[0][i] != 0 && board[0][i] == board[1][i] && board[1][i] == board[2][i]) {
                return board[0][i] == 1 ? "A" : "B";
            }
        }
        if (board[0][0] != 0 && board[0][0] == board[1][1] && board[1][1] == board[2][2]) {
            return board[0][0] == 1 ? "A" : "B";
        }
        if (board[0][2] != 0 && board[0][2] == board[1][1] && board[1][1] == board[2][0]) {
            return board[0][2] == 1 ? "A" : "B";
        }
        return "Pending";
    }
}