<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 absent, int consecutiveLate, vector<vector<vector<int>>> &dp) {
        if (n < 0 || absent < 0 || consecutiveLate >= 3) {
            return 0;
        }

        if (n == 0) {
            return 1;
        }

        if (dp[n][absent][consecutiveLate] != -1) {
            return dp[n][absent][consecutiveLate];
        }

        int ans = 0;
        // Case 1 : Present ('P') on nth Day
        ans = (ans + solve(n - 1, absent, 0, dp) % mod) % mod;

        // Case 1 : Absent ('A') on nth Day
        ans = (ans + solve(n - 1, absent - 1, 0, dp) % mod) % mod;

        // Case 1 : Late ('L') on nth Day
        ans = (ans + solve(n - 1, absent, consecutiveLate + 1, dp) % mod) % mod;

        return dp[n][absent][consecutiveLate] = ans;
    }
    int checkRecord(int n) {
        // vector<vector<vector<int>>> dp (n + 1, vector<vector<int>> (2, vector<int> (3, -1)));
        
        // return solve(n, 1, 0, dp);

        vector<vector<vector<int>>> dp (n + 1, vector<vector<int>> (2, vector<int> (4, 0)));
        for (int i = 0; i <= n; i += 1) {
            for (int absent = 0; absent <= 1; absent += 1) {
                for (int consecutiveLate = 0; consecutiveLate <= 2; consecutiveLate += 1) {
                    if (i == 0) {
                        dp[i][absent][consecutiveLate] = 1;
                        continue;
                    }
                    int ans = 0;
                    // Case 1 : Present ('P') on nth Day
                    ans = (ans + dp[i - 1][absent][0] % mod) % mod;

                    // Case 1 : Absent ('A') on nth Day
                    if (absent - 1 >= 0) {
                        ans = (ans + dp[i - 1][absent - 1][0] % mod) % mod;
                    } 

                    // Case 1 : Late ('L') on nth Day
                    ans = (ans + dp[i - 1][absent][consecutiveLate + 1] % mod) % mod;

                    dp[i][absent][consecutiveLate] = ans;
                }
            }
        }

        return dp[n][1][0];
    }
};
275134