<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>

#include <vector>
using namespace std;

class Solution {
public:
    int N;
    int mod = 1000000007;
    vector<vector<vector<vector<int>>>> memo;

    int f(int L, int A, int prev, int i) {
        if (i == N) {
            return 1; // We've constructed a valid sequence
        }
        if (memo[L][A][prev][i] != -1) {
            return memo[L][A][prev][i];
        }
        int ans = 0;

        // Case 1: Add 'P' (Present)
        ans = (ans + f(0, A, 0, i + 1)) % mod;

        // Case 2: Add 'L' (Late)
        if (L < 2) {
            ans = (ans + f(L + 1, A, 1, i + 1)) % mod;
        }

        // Case 3: Add 'A' (Absent)
        if (A == 0) {
            ans = (ans + f(0, A + 1, 2, i + 1)) % mod;
        }

        return memo[L][A][prev][i] = ans;
    }

    int checkRecord(int n) {
        N = n;
        memo = vector<vector<vector<vector<int>>>>(3, vector<vector<vector<int>>>(2, vector<vector<int>>(3, vector<int>(n + 1, -1))));
        return f(0, 0, 0, 0);
    }
};

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