<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 {
private:
    int dfs(int len, int ab, int pp, int p, int n, vector<vector<vector<vector<int>>>>& dp) {
        if (len == n) return 1;
        if (dp[len][ab][pp][p] != -1) return dp[len][ab][pp][p];
        
        int pres = dfs(len + 1, ab, 0, 0, n, dp);
        int abs = 0;
        if (ab + 1 < 2) {
            abs = dfs(len + 1, ab + 1, 0, 0, n, dp);
        }
        int late = 0;
        if (!(pp && p)) {
            late = dfs(len + 1, ab, 1, pp, n, dp);
        }

        return dp[len][ab][pp][p] = ((pres + abs)  % 1000000007 + late)  % 1000000007; 
    }

public:
    int checkRecord(int n) {
        vector<vector<vector<vector<int>>>> dp(n + 1, vector<vector<vector<int>>>(2, vector<vector<int>>(2, vector<int>(2, -1))));
        return dfs(0, 0, 0, 0, n, dp);
    }
};

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