<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:
        MOD = 1000000007
        # Initialize the cache to store sub-problem results.
        memo = [[[-1] * 3 for _ in range(2)] for _ in range(n + 1)]

        # Recursive function to return the count of combinations
        # of length 'n' eligible for the award.
        def eligible_combinations(n, total_absences, consecutive_lates):
            # If the combination has become not eligible for the award,
            # then we will not count any combinations that can be made using it.
            if total_absences >= 2 or consecutive_lates >= 3:
                return 0
            # If we have generated a combination of length 'n' we will count it.
            if n == 0:
                return 1
            # If we have already seen this sub-problem earlier,
            # we return the stored result.
            if memo[n][total_absences][consecutive_lates] != -1:
                return memo[n][total_absences][consecutive_lates]

            # We choose 'P' for the current position.
            count = eligible_combinations(n - 1, total_absences, 0)
            # We choose 'A' for the current position.
            count = (
                count + eligible_combinations(n - 1, total_absences + 1, 0)
            ) % MOD
            # We choose 'L' for the current position.
            count = (
                count
                + eligible_combinations(
                    n - 1, total_absences, consecutive_lates + 1
                )
            ) % MOD

            # Return and store the current sub-problem result in the cache.
            memo[n][total_absences][consecutive_lates] = count
            return count

        # Return count of combinations of length 'n' eligible for the award.
        return eligible_combinations(n, 0, 0)
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