1275. Find Winner on a Tic Tac Toe Game

Description

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.

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.

Approach

Links

• Binarysearch

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: moves = [[0, 0], [2, 0], [1, 1], [2, 1], [2, 2]]

Output: "A"

Explanation: "A" wins, he always plays 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.

"XXO"

"OOX"

"XOX"

Example 4

Input: moves = [[0,0],[1,1]]

Output: "Pending"

Explanation: The game has not finished yet.

"X "

" O "

" "

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    
    public String tictactoe(int[][] moves) {
        if(moves.length > 4) {
            int n = 3;
            char[] players = {'A', 'B'};
            char[][] board = new char[n][n];
            for(int i = 0; i < moves.length; i++) {
                int[] move = moves[i];
                board[move[0]][move[1]] = players[i%2];
            }
        
            for(int i = 0; i < n; i++) {
                for(char player: players) {
                    if(rowCheck(board, i, player) || colCheck(board, i, player)) {
                        return player + "";
                    }
                }
            }

            for(char player: players) {
                if(diagonalCheck(board, player)) {
                    return player + "";
                }
            }
        }
        
        return moves.length == 9? "Draw": "Pending";
    }
    
    private boolean rowCheck(char[][] board, int row, char player) {
        return board[row][0] == player && board[row][1] == player && board[row][2] == player;
    }
    
    private boolean colCheck(char[][] board, int col, char player) {
        return board[0][col] == player && board[1][col] == player && board[2][col] == player;
    }
    
    private boolean diagonalCheck(char[][] board, char player) {
        return (board[0][0] == player && board[1][1] == player && board[2][2] == player) || 
            (board[2][0] == player && board[1][1] == player && board[0][2] == player);
    }
}

Solution 2

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    private final int n = 3;
    
    public String tictactoe(int[][] moves) {
        if(moves.length > 4) {
            int[] rows = new int[n];
            int[] cols = new int[n];
            int diagonal = 0, antiDiagonal = 0;
            
            int player = 1;
            
            for(int[] move: moves) {
                int x = move[0], y = move[1];
                
                if(x == y) {
                    diagonal += player;
                }
                if(x+y == n-1) {
                    antiDiagonal += player;
                }
                
                rows[x] += player;
                cols[y] += player;
                
                if(Math.abs(rows[x]) == n || Math.abs(cols[y]) == n ||
                   Math.abs(diagonal) == n || Math.abs(antiDiagonal) == n) {
                    return (player == 1)? "A": "B";
                }
                
                player *= -1;
            }
        }
        
        return moves.length == n*n? "Draw": "Pending";
    }
}

Follow up