<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&#39;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>&quot;Alice&quot;</code><em> if Alice will win, </em><code>&quot;Bob&quot;</code><em> if Bob will win, or </em><code>&quot;Tie&quot;</code><em> if they will end the game with the same score</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> stoneValue = [1,2,3,7]
<strong>Output:</strong> &quot;Bob&quot;
<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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> stoneValue = [1,2,3,-9]
<strong>Output:</strong> &quot;Alice&quot;
<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&#39;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&#39;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.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> stoneValue = [1,2,3,6]
<strong>Output:</strong> &quot;Tie&quot;
<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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= stoneValue.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-1000 &lt;= stoneValue[i] &lt;= 1000</code></li>
</ul>

class Solution {
public:
     int t[50005][2];
    int sol(int i,  const vector<int>& p, int isAlice) {
        if (i >= p.size()) return 0;

        if(t[i][isAlice]!=-1) return t[i][isAlice];

        int ans = isAlice ? INT_MIN : INT_MAX;

        for (int x = 1; x <= 3 && i + x <= p.size(); ++x) {
            int temp = p[i + x - 1];
            if (i > 0) temp -= p[i - 1];
            
            if (isAlice) {
                ans = max(ans, temp + sol(i + x,  p, 0));
            } else {
                ans = min(ans, sol(i + x,  p, 1));
            }
        }

        return t[i][isAlice] = ans;
    }
    string stoneGameIII(vector<int>& stoneValue) {
        int n = stoneValue.size();
        vector<int> p(n);
       memset(t,-1,sizeof(t));
        // Create prefix sum array
        p[0] = stoneValue[0];
        for (int i = 1; i < n; ++i) {
            p[i] = p[i - 1] + stoneValue[i];
        }
        int total=accumulate(stoneValue.begin(),stoneValue.end(),0LL);
        int Alice= sol(0, p, 1);
        if((total-Alice)>Alice) return "Bob";
        else if((total-Alice)<Alice) return "Alice";
        return "Tie";
    }
};
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