<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==1:
            return 3
        dl= [[0,0]for _ in range(n)] #no of records ending with two 'L's, 1: no absent, 0: one absent
        sl= [[0,0]for _ in range(n)] #no of records ending with one 'L', 1: no absent, 0: one absent
        nl= [[0,0]for _ in range(n)] #no of records with no 'L' at the end, 1: no absent, 0: one absent 
        dl[1][1]= 1
        sl[1][0], sl[1][1]= 1,1
        nl[1][0], nl[1][1]= 3,2   #Initializations
        k= 10**9 + 7
        for i in range(2,n):
            sl[i][0],sl[i][1]= nl[i-1][0],nl[i-1][1]
            nl[i][0]= (dl[i-1][0]+ dl[i-1][1] + sl[i-1][0] + sl[i-1][1] + nl[i-1][0] + nl[i-1][1])%k
            nl[i][1]= (nl[i-1][1]+ sl[i-1][1] + dl[i-1][1])%k
            dl[i][0],dl[i][1]= sl[i-1][0],sl[i-1][1]
        return (sum(dl[-1])+sum(sl[-1])+sum(nl[-1]))%k
45588