Design a Tic-tac-toe game that is played between two players on a n x n grid.
You may assume the following rules:
- A move is guaranteed to be valid and is placed on an empty block.
- Once a winning condition is reached, no more moves is allowed.
- A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.
Example:
Given n = 3, assume that player 1 is "X" and player 2 is "O" in the board. TicTacToe toe = new TicTacToe(3); toe.move(0, 0, 1); -> Returns 0 (no one wins) |X| | | | | | | // Player 1 makes a move at (0, 0). | | | | toe.move(0, 2, 2); -> Returns 0 (no one wins) |X| |O| | | | | // Player 2 makes a move at (0, 2). | | | | toe.move(2, 2, 1); -> Returns 0 (no one wins) |X| |O| | | | | // Player 1 makes a move at (2, 2). | | |X| toe.move(1, 1, 2); -> Returns 0 (no one wins) |X| |O| | |O| | // Player 2 makes a move at (1, 1). | | |X| toe.move(2, 0, 1); -> Returns 0 (no one wins) |X| |O| | |O| | // Player 1 makes a move at (2, 0). |X| |X| toe.move(1, 0, 2); -> Returns 0 (no one wins) |X| |O| |O|O| | // Player 2 makes a move at (1, 0). |X| |X| toe.move(2, 1, 1); -> Returns 1 (player 1 wins) |X| |O| |O|O| | // Player 1 makes a move at (2, 1). |X|X|X|
Follow up:
Could you do better than O(n2) per
Could you do better than O(n2) per
move()
operation?
Hint:
- Could you trade extra space such that
move()
operation can be done in O(1)? - You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.
Use addtional arrays rows[n], cols[n] and two varialbes diagonal, anti_diagonal to mark the number of Xs and Os.
Code (Java):
public class TicTacToe { private int[][] rows; private int[][] cols; private int[] diag; private int[] xdiag; private int n; /** Initialize your data structure here. */ public TicTacToe(int n) { this.n = n; rows = new int[2][n]; cols = new int[2][n]; diag = new int[2]; xdiag = new int[2]; } /** Player {player} makes a move at ({row}, {col}). @param row The row of the board. @param col The column of the board. @param player The player, can be either 1 or 2. @return The current winning condition, can be either: 0: No one wins. 1: Player 1 wins. 2: Player 2 wins. */ public int move(int row, int col, int player) { int p = player == 1 ? 0 : 1; rows[p][row]++; cols[p][col]++; if (row == col) { diag[p]++; } // X-diagonal if (row + col == n - 1) { xdiag[p]++; } // If any of them equals to n, return 1 if (rows[p][row] == n || cols[p][col] == n || diag[p] == n || xdiag[p] == n) { return p + 1; } return 0; } } /** * Your TicTacToe object will be instantiated and called as such: * TicTacToe obj = new TicTacToe(n); * int param_1 = obj.move(row,col,player); */
A Neat Code:
In the previous solution, we allocate two arrays for player 1 and 2, respectively. Actually we can use only one array for both of the players. Say, if it is player 1 put one chess, add that location by 1. If it is player 2, deduce it by one. Finally, if either player 1 or player 2 win, that location must be equal to n or -n.
Code (Java):
public class TicTacToe { private int[] rows; private int[] cols; private int diag; private int xdiag; private int n; /** Initialize your data structure here. */ public TicTacToe(int n) { this.n = n; rows = new int[n]; cols = new int[n]; diag = 0; xdiag = 0; } /** Player {player} makes a move at ({row}, {col}). @param row The row of the board. @param col The column of the board. @param player The player, can be either 1 or 2. @return The current winning condition, can be either: 0: No one wins. 1: Player 1 wins. 2: Player 2 wins. */ public int move(int row, int col, int player) { int count = player == 1 ? 1 : -1; rows[row] += count; cols[col] += count; if (row == col) { diag += count; } // X-diagonal if (row + col == n - 1) { xdiag += count; } // If any of them equals to n, return 1 if (Math.abs(rows[row]) == n || Math.abs(cols[col]) == n || Math.abs(diag) == n || Math.abs(xdiag) == n) { return count > 0 ? 1 : 2; } return 0; } } /** * Your TicTacToe object will be instantiated and called as such: * TicTacToe obj = new TicTacToe(n); * int param_1 = obj.move(row,col,player); */
Hey Guys, I took a mock interview of a software engineer at Paypal. You can checkout the approach :)
ReplyDeletehttps://www.youtube.com/watch?v=see8KokFDl8
My channel link
https://www.youtube.com/channel/UCZacWEbniiEaybfVdcuzAPw