<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:
    int mod=1e9+7;
    int solve(int n,int abs,int late, vector<vector<vector<int>>>&dp)
    {
        if(n==0)
            return 1;
        int ans=0;
        if(dp[n][abs][late]!=-1)
            return dp[n][abs][late];

        if (abs < 1) 
        ans = (ans + solve(n - 1, abs + 1, 0, dp)) % mod;
        if(late<2)
        ans = (ans + solve(n - 1, abs, late + 1, dp)) % mod;
        ans = (ans + solve(n - 1, abs, 0, dp)) % mod;
        return dp[n][abs][late]=ans;
        
    }
    int checkRecord(int n) {
    vector<vector<int>>prev(2,vector<int>(3,0));
    vector<vector<int>>next(2,vector<int>(3,0));
     prev[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
        next = vector<vector<int>>(2, vector<int>(3, 0));
        for (int a = 0; a <= 1; ++a) {
            for (int l = 0; l <= 2; ++l) {
                next[a][0] = (next[a][0] + prev[a][l]) % mod;
                if (l < 2) {
                    next[a][l + 1] = (next[a][l + 1] + prev[a][l]) % mod;
                }
                if (a < 1) {
                    next[a + 1][0] = (next[a + 1][0] + prev[a][l]) % mod;
                }
            }
        }
        prev=next;
    }

    int res = 0;
    for (int a = 0; a <= 1; ++a) {
        for (int l = 0; l <= 2; ++l) {
            res = (res + next[a][l]) % mod;
        }
    }

    return res;
}


};
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