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.
class Solution {
public String tictactoe(int[][] moves) {
int[] arow = new int[3], brow = new int[3], acol = new int[3], bcol = new int[3];
int aD1 = 0, bD1 = 0, aD2 = 0, bD2 = 0;
for(int i = 0; i < moves.length; i++){
int r = moves[i][0], c = moves[i][1];
if(i % 2 == 0){
if(++arow[r] == 3 || ++acol[c] == 3 || r==c && ++aD1 == 3 || r + c == 2 && ++aD2 == 3) return "A";
}
else if(++brow[r] == 3 || ++bcol[c] == 3 || r==c && ++bD1 == 3 || r + c == 2 && ++bD2 == 3) return "B";
}
return moves.length == 9 ? "Draw" : "Pending";
}
}