# Design Tic-Tac-Toe

`Design`

Medium

Design a Tic-tac-toe game that is played between two players on anxngrid.

You may assume the following rules:

1. A move is guaranteed to be valid and is placed on an empty block.
2. Once a winning condition is reached, no more moves is allowed.
3. 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|
``````

Could you do better than O(n^2) per`move()`operation?

Could you get O(1) per `move()` operation?

## Solution & Analysis

#### O(n) Time, O(n^2) Space solution

``````class TicTacToe {
private char[][] board;
private static char X = 'X';
private static char O = 'O';
private int size;

/** Initialize your data structure here. */
public TicTacToe(int n) {
board = new char[n][n];
size = n;
}

/** 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) {
char c;
if (player == 1) {
c = X;
} else {
c = O;
}
if (board[row][col] != 0) {
// throw error, occupied
return 0;
}
board[row][col] = c;

if (hasWon(row, col, size, c)) {
return player;
}
return 0;
}

private boolean hasWon(int row, int col, int n, char c) {

// check horizontal
boolean rowLine = true;
for (int i = 0; i < n; i++) {
rowLine = rowLine && (board[i][col] == c);
}
// check vertical
boolean colLine = true;
for (int j = 0; j < n; j++) {
colLine = colLine && (board[row][j] == c);
}
// check diagonal
if (row + col == n - 1 || row == col) {
boolean diagLine1 = true;
boolean diagLine2 = true;
for (int j = 0; j < n; j++) {
diagLine1 = diagLine1 && (board[j][j] == c);
}
for (int j = 0; j < n; j++) {
diagLine2 = diagLine2 && (board[n - 1 - j][j] == c);
}
return rowLine || colLine || diagLine1 || diagLine2;
} else {
return rowLine || colLine;
}
}
}

/**
* Your TicTacToe object will be instantiated and called as such:
* TicTacToe obj = new TicTacToe(n);
* int param_1 = obj.move(row,col,player);
*/
``````

#### O(1) Time, O(n) Space solution

The key observation is that in order to win Tic-Tac-Toe you must have the entire row or column. Thus, we don't need to keep track of an entire n^2 board. We only need to keep a count for each row and column. If at any time a row or column matches the size of the board then that player has won.

To keep track of which player, I add one for Player1 and -1 for Player2. There are two additional variables to keep track of the count of the diagonals. Each time a player places a piece we just need to check the count of that row, column, diagonal and anti-diagonal.

``````public class TicTacToe {
private int[] rows;
private int[] cols;
private int diagonal;
private int antiDiagonal;

/** Initialize your data structure here. */
public TicTacToe(int n) {
rows = new int[n];
cols = new int[n];
}

/** 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 toAdd = player == 1 ? 1 : -1;

if (row == col) {
}

if (col + row == cols.length - 1) {
}

int size = rows.length;
if (Math.abs(rows[row]) == size ||
Math.abs(cols[col]) == size ||
Math.abs(diagonal) == size  ||
Math.abs(antiDiagonal) == size) {
return player;
}

return 0;
}
}
``````

Not using Math.abs(), changing matching target @hot13399:

``````class TicTacToe {

int[] rows, cols;
int d1, d2, n;

public TicTacToe(int n) {
rows = new int[n];
cols = new int[n];
d1 = 0;
d2 = 0;
this.n = n;
}

public int move(int row, int col, int player) {
int val = (player == 1) ? 1 : -1;
int target = (player == 1) ? n : -n;

if(row == col) { // diagonal
d1 += val;
if(d1 == target) return player;
}

if(row + col + 1 == n) { // diagonal
d2 += val;
if(d2 == target) return player;
}

rows[row] += val;
cols[col] += val;

if(rows[row] == target || cols[col] == target) return player;

return 0;
}
}
``````

## Reference

https://leetcode.com/problems/design-tic-tac-toe/discuss/81898/Java-O(1)-solution-easy-to-understand