<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>

#define MOD 1000000007
class Solution {
public:
    vector<vector<vector<int>>> memo;  // 3D vector for memoization

    int solve(int idx, int n, int absentCnt, int lateCnt) {
        // Base conditions
        if (absentCnt >= 2 || lateCnt >= 3) return 0;  // Invalid record
        if (idx == n) return 1;  // Reached the end of the record and it's valid
        
        // If this state has already been computed, return the memoized result
        if (memo[idx][absentCnt][lateCnt] != -1) {
            return memo[idx][absentCnt][lateCnt];
        }

        // Recur for each possible character ('A', 'L', 'P')
        int result = 0;

        // Option 1: Add 'A' (absent), reset late count
        result += solve(idx + 1, n, absentCnt + 1, 0);
        result = result%MOD;
        // Option 2: Add 'L' (late), increment late count
        result += solve(idx + 1, n, absentCnt, lateCnt + 1);
        result = result%MOD;
        // Option 3: Add 'P' (present), reset late count
        result += solve(idx + 1, n, absentCnt, 0);
result = result%MOD;
        // Memoize the result for the current state
        memo[idx][absentCnt][lateCnt] = result%MOD;
        return result%MOD;
    }

    int checkRecord(int n) {
        // Initialize the 3D vector with dimensions [n+1][2][3], filled with -1 (uncomputed state)
        memo = vector<vector<vector<int>>>(n + 1, vector<vector<int>>(2, vector<int>(3, -1)));

        // Start with index 0, 0 absences, and 0 late days
        return solve(0, n, 0, 0)%MOD;
    }
};

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