<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>

// dp(i, flagLate, flagAbsent, lateConsecutive, flag)
int dp[100100][5][10][2];
int n;
int mod = 1e9+7;
int rec(int i, int flagAbsent, int lateConsecutive, int flag){
    // pruning
    if(flagAbsent>=2 || lateConsecutive>=3 || flag==1)
        return 0;
    // base case
    if(i==n){
        if(flagAbsent<2 && lateConsecutive<3 && flag==0)
            return 1;
        else return 0;
    }
    // cache
    if(dp[i][flagAbsent][lateConsecutive][flag]!=-1)
        return dp[i][flagAbsent][lateConsecutive][flag];
    // compute
    int ans =0;
    if(lateConsecutive+1>=3)ans = (ans + rec(i+1,flagAbsent, lateConsecutive+1, 1)%mod)%mod; //late
    else if(lateConsecutive+1<3) ans = (ans + rec(i+1,flagAbsent, lateConsecutive+1, flag)%mod)%mod; //late
    ans = (ans + rec(i+1, flagAbsent+1, 0, flag)%mod)%mod; //absent
    ans = (ans + rec(i+1, flagAbsent, 0, flag)%mod)%mod; //present
    // save and return
    return dp[i][flagAbsent][lateConsecutive][flag]= ans;
}


class Solution {
public:
    int checkRecord(int sz) {
        n= sz;
        memset(dp, -1, sizeof(dp));
        return rec(0,0,0,0);
    }
};
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