<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>

int dp[100005][27][4][3];
int mod = 1e9 + 7;
#define ll long long int
class Solution {
public:
    
    int solve(int i, int prev, int prevCount, int totalAbsent, int n){
        // Base case:-1
        if(prevCount>=3 and prev=='L'-'A'){
            return 0;
        }
        
        // Base case:- 2
        if(totalAbsent>=2){
            return 0;
        }
        
        // Base Case:-3
        if(i>=n){
            return 1; // if we reach to the end we have one possible solution;
        }
        
        if(dp[i][prev][prevCount][totalAbsent]!=-1) return dp[i][prev][prevCount][totalAbsent];
        
        // at every point we have 3 options;
        ll absent = (prev=='A'-'A'?solve(i+1,prev,min(prevCount+1,3),min(totalAbsent+1,2),n):solve(i+1,'A'-'A',1,min(totalAbsent+1,2),n));
        ll late = (prev=='L'-'A'?solve(i+1,prev,min(prevCount+1,3),totalAbsent,n):solve(i+1,'L'-'A',1,totalAbsent,n));
        ll present = (prev=='P'-'A'?solve(i+1,prev,min(prevCount+1,3),totalAbsent,n):solve(i+1,'P'-'A',1,totalAbsent,n));
        return dp[i][prev][prevCount][totalAbsent] = (absent%mod + late%mod + present%mod)%mod;
        
    }
    
    int checkRecord(int n) {
        memset(dp,-1,sizeof(dp));
        return solve(0,26,0,0,n);
    }
};
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