# [Programming Problem] Design Tic-Tac-Toe

Assume the following rules are for the tic-tac-toe game on an n x n board between two players:

– A move is guaranteed to be valid and is placed on an empty block.
– Once a winning condition is reached, no more moves are allowed.
– A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.

Implement the TicTacToe class:

– TicTacToe(int n) Initializes the object the size of the board n.
– int move(int row, int col, int player) Indicates that player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move.

Follow up:
Could you do better than O(n2) per move() operation?

Key here is that every move is going to be valid, so we don’t need to check for invalid or repeat moves! Instead of keeping track of the whole board and checking for winners we can simply keep track of number of times a player has hit a row, col, or either of the diagonals. If count for any of them for a player hits ‘N’ we have a winner!

Few things to keep in mind:-

• Keep player status (i.e. hits for rows, cols, and both diagonals) for both players. A single object wont do.
• Remember we have 2 diagonals. When calculating hits for diagonals, treat those diagonals as separate variables! (not single diagonal)
```/** * Initialize your data structure here. * @param {number} n */ var TicTacToe = function(n) {   // - Gotcha 1: Please add player status for both players, single object wont do // We need to keep track if specifically one players rows, cols, or diagonals have hit 'N' // - Gotcha 2: Please remember we have 2 diagonals! // - Gotcha 3:   // Player status objects! this.playerOne = { rows: new Array(n).fill(0), cols: new Array(n).fill(0), diagonal: [0, 0], }   this.playerTwo = { rows: new Array(n).fill(0), cols: new Array(n).fill(0), diagonal: [0, 0], }   // N this.n = n;   // rows, cols in diagonal one (and diagonal two) this.diagonal_one_set = new Set(); this.diagonal_two_set = new Set();   // diagonal 1 and diagonal 2 let i = this.n-1; let j = 0; while (i >=0 && j < this.n ) { this.diagonal_two_set.add(i + '' + j); i--; j++; }   i = 0; j = 0; while (i < this.n && j < this.n ) { this.diagonal_one_set.add(i + '' + j); i++; j++; } };   /** * 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. * @param {number} row * @param {number} col * @param {number} player * @return {number} */ TicTacToe.prototype.move = function(row, col, player) { const playerStatus = player === 1 ? this.playerOne : this.playerTwo;   playerStatus.rows[row]++; playerStatus.cols[col]++; if (this.diagonal_one_set.has(row + '' + col)) playerStatus.diagonal[0]++; if (this.diagonal_two_set.has(row + '' + col)) playerStatus.diagonal[1]++;   if ( playerStatus.rows[row] === this.n || playerStatus.cols[col] === this.n || playerStatus.diagonal[0] === this.n || playerStatus.diagonal[1] === this.n) { return player; } else { return 0; } };   /** * Your TicTacToe object will be instantiated and called as such: * var obj = new TicTacToe(n) * var param_1 = obj.move(row,col,player) */```

### 32 thoughts on “[Programming Problem] Design Tic-Tac-Toe”

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