Dice Roll Simulation
<p>A die simulator generates a random number from <code>1</code> to <code>6</code> for each roll. You introduced a constraint to the generator such that it cannot roll the number <code>i</code> more than <code>rollMax[i]</code> (<strong>1-indexed</strong>) consecutive times.</p>

<p>Given an array of integers <code>rollMax</code> and an integer <code>n</code>, return <em>the number of distinct sequences that can be obtained with exact </em><code>n</code><em> rolls</em>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Two sequences are considered different if at least one element differs from each other.</p>

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

<pre>
<strong>Input:</strong> n = 2, rollMax = [1,1,2,2,2,3]
<strong>Output:</strong> 34
<strong>Explanation:</strong> There will be 2 rolls of die, if there are no constraints on the die, there are 6 * 6 = 36 possible combinations. In this case, looking at rollMax array, the numbers 1 and 2 appear at most once consecutively, therefore sequences (1,1) and (2,2) cannot occur, so the final answer is 36-2 = 34.
</pre>

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

<pre>
<strong>Input:</strong> n = 2, rollMax = [1,1,1,1,1,1]
<strong>Output:</strong> 30
</pre>

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

<pre>
<strong>Input:</strong> n = 3, rollMax = [1,1,1,2,2,3]
<strong>Output:</strong> 181
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 5000</code></li>
	<li><code>rollMax.length == 6</code></li>
	<li><code>1 &lt;= rollMax[i] &lt;= 15</code></li>
</ul>
Array; Dynamic Programming
class Solution:
    def dieSimulator(self, n: int, rollMax: List[int]) -> int:
        @cache
        def dfs(i, j, x):
            if i >= n:
                return 1
            ans = 0
            for k in range(1, 7):
                if k != j:
                    ans += dfs(i + 1, k, 1)
                elif x < rollMax[j - 1]:
                    ans += dfs(i + 1, j, x + 1)
            return ans % (10**9 + 7)

        return dfs(0, 0, 0)
