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JavaScript Object-Oriented JavaScript: Challenge Adding the Game Logic checkForWin() Method Challenge Solution

Posted by matthewchau2
matthewchau2
8,301 Points

Is my solution also acceptable?

I understand how Ashley did it, but my code seems to work as well. We just went about it a little differently for the horizontal and diagonal directions. Just want to confirm that my solution is also acceptable.

/**
     * checks if there is a winner on the board after each token drop
     * @param   {object}   target - targeted space for dropped token  
     * @return  {boolean}  boolean value indicating whether the game has been won (true) or not (false) 
     */
    checkForWin(target) {
        const owner = target.token.owner;
        let win = false;

        // vertical
        for (let x = 0; x < this.board.columns; x++) {
            for (let y = 0; y < this.board.rows - 3; y++) {
                if (this.board.spaces[x][y].owner === owner &&
                    this.board.spaces[x][y + 1].owner === owner &&
                    this.board.spaces[x][y + 2].owner === owner &&
                    this.board.spaces[x][y + 3].owner === owner) {
                    win = true;    
                }
            }
        }

        // horizontal
        for (let y = 0; y < this.board.rows; y++) {
            for (let x = 0; x < this.board.columns - 3; x++) {
                if (this.board.spaces[x][y].owner === owner &&
                    this.board.spaces[x + 1][y].owner === owner &&
                    this.board.spaces[x + 2][y].owner === owner &&
                    this.board.spaces[x + 3][y].owner === owner) {
                    win = true;
                }
            }
        }

        // diagonal downwardly left to right
        for (let x = 0; x < this.board.columns - 3; x++) {
            for (let y = 0; y < this.board.rows - 3; y++) {
                if (this.board.spaces[x][y].owner === owner &&
                    this.board.spaces[x + 1][y + 1].owner === owner &&
                    this.board.spaces[x + 2][y + 2].owner === owner &&
                    this.board.spaces[x + 3][y + 3].owner === owner) {
                    win = true;
                }
            }
        }

        // diagonal downwardly right to left
        for (let x = 6; x > this.board.columns - 5; x--) {
            for (let y = 0; y < this.board.rows - 3; y++) {
                if (this.board.spaces[x][y].owner === owner &&
                    this.board.spaces[x - 1][y + 1].owner === owner &&
                    this.board.spaces[x - 2][y + 2].owner === owner &&
                    this.board.spaces[x - 3][y + 3].owner === owner) {
                    win = true;
                }
            }
        }

        return win;
    }

1 Answer

Steven Parker
Steven Parker
229,783 Points
Steven Parker
Steven Parker
229,783 Points

In programming, there is rarely just one way to create a solution. The more complex the problem, the larger the opportunity that different but equally valid solutions can be created.

And in this case, the order in which the possible win lines are examined is completely unimportant as long as they all get checked.