<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 {
public:
    const int mod = 1e9+7;
    int mod_add(int a, int b) {
        return (a%mod + b%mod)%mod;
    }

    vector<vector<vector<int> > > dp;

    int findPossibleRecords(int n, int idx, int absentCnt, int lateCnt) {
        if(idx == n)
            return 1;

        if(dp[idx][absentCnt][lateCnt] != -1)
            return dp[idx][absentCnt][lateCnt];

        int ans = 0;

        // present at current day 
        ans = mod_add(ans, findPossibleRecords(n, idx+1, absentCnt, 0));
        
        if(absentCnt < 1) {
            // absent today 
            ans = mod_add(ans, findPossibleRecords(n, idx+1, absentCnt+1, 0));
        }

        // late today 
        if(lateCnt < 2)
            ans = mod_add(ans, findPossibleRecords(n, idx+1, absentCnt, lateCnt+1));

        return dp[idx][absentCnt][lateCnt] = ans;
    }

    int checkRecord(int n) {
        
        dp.assign(n, vector<vector<int> > (2, vector<int> (3, -1)));
        int ans = findPossibleRecords(n, 0, 0, 0);
        return ans;
    }
};
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