<p>An attendance record for a student can be represented as a string where each character signifies whether the student was absent, late, or present on that day. The record only contains the following three characters:</p>

<ul>
	<li><code>&#39;A&#39;</code>: Absent.</li>
	<li><code>&#39;L&#39;</code>: Late.</li>
	<li><code>&#39;P&#39;</code>: Present.</li>
</ul>

<p>Any student is eligible for an attendance award if they meet <strong>both</strong> of the following criteria:</p>

<ul>
	<li>The student was absent (<code>&#39;A&#39;</code>) for <strong>strictly</strong> fewer than 2 days <strong>total</strong>.</li>
	<li>The student was <strong>never</strong> late (<code>&#39;L&#39;</code>) for 3 or more <strong>consecutive</strong> days.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the <strong>number</strong> of possible attendance records of length</em> <code>n</code><em> that make a student eligible for an attendance award. The answer may be very large, so return it <strong>modulo</strong> </em><code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 8
<strong>Explanation:</strong> There are 8 records with length 2 that are eligible for an award:
&quot;PP&quot;, &quot;AP&quot;, &quot;PA&quot;, &quot;LP&quot;, &quot;PL&quot;, &quot;AL&quot;, &quot;LA&quot;, &quot;LL&quot;
Only &quot;AA&quot; is not eligible because there are 2 absences (there need to be fewer than 2).
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 3
</pre>

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

<pre>
<strong>Input:</strong> n = 10101
<strong>Output:</strong> 183236316
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def checkRecord(self, n: int) -> int:
        if n < 0:
            raise ValueError("n must be non-negative")
        if n == 0:
            return 0

        MOD = 10**9 + 7

        # Initialize the DP array
        # dp[i][j][k] represents the number of valid sequences of length i,
        # ending with j consecutive 'L's, and having k total 'A's
        dp = [[[0 for _ in range(2)] for _ in range(3)] for _ in range(n + 1)]

        # Base case: sequence of length 1
        dp[1][0][0] = 1  # 'P'
        dp[1][1][0] = 1  # 'L'
        dp[1][0][1] = 1  # 'A'

        # Fill the DP array
        for i in range(2, n + 1):
            for j in range(3):  # j represents the number of consecutive 'L's ending at i
                for k in range(2):  # k represents the number of 'A's in the sequence
                    if j == 0:  # Ending with 'P' or 'A'
                        # Can append 'P' to any previous state
                        dp[i][0][k] = (dp[i][0][k] + sum(dp[i-1][l][k] for l in range(3))) % MOD
                        if k > 0:
                            # Can append 'A' to any previous state without 'A'
                            dp[i][0][k] = (dp[i][0][k] + sum(dp[i-1][l][k-1] for l in range(3))) % MOD
                    else:  # Ending with 'L'
                        # Can only append 'L' if there were less than 2 'L's before
                        dp[i][j][k] = dp[i-1][j-1][k]

        # Sum up all valid sequences of length n
        result = sum(dp[n][j][k] for j in range(3) for k in range(2)) % MOD

        return result
58906