<p>Alice and Bob continue their games with piles of stones. There are several stones <strong>arranged in a row</strong>, and each stone has an associated value which is an integer given in the array <code>stoneValue</code>.</p>
<p>Alice and Bob take turns, with Alice starting first. On each player's turn, that player can take <code>1</code>, <code>2</code>, or <code>3</code> stones from the <strong>first</strong> remaining stones in the row.</p>
<p>The score of each player is the sum of the values of the stones taken. The score of each player is <code>0</code> initially.</p>
<p>The objective of the game is to end with the highest score, and the winner is the player with the highest score and there could be a tie. The game continues until all the stones have been taken.</p>
<p>Assume Alice and Bob <strong>play optimally</strong>.</p>
<p>Return <code>"Alice"</code><em> if Alice will win, </em><code>"Bob"</code><em> if Bob will win, or </em><code>"Tie"</code><em> if they will end the game with the same score</em>.</p>
<strong>Explanation:</strong> Alice will always lose. Her best move will be to take three piles and the score become 6. Now the score of Bob is 7 and Bob wins.
<strong>Explanation:</strong> Alice must choose all the three piles at the first move to win and leave Bob with negative score.
If Alice chooses one pile her score will be 1 and the next move Bob's score becomes 5. In the next move, Alice will take the pile with value = -9 and lose.
If Alice chooses two piles her score will be 3 and the next move Bob's score becomes 3. In the next move, Alice will take the pile with value = -9 and also lose.
Remember that both play optimally so here Alice will choose the scenario that makes her win.
<strong>Explanation:</strong> Alice cannot win this game. She can end the game in a draw if she decided to choose all the first three piles, otherwise she will lose.
