<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 static int MOD = 1000000007;
    int dfs(int n, int day, int absences, int lates, vector<vector<vector<int>>>& memo) {
        if (absences == 2 || lates == 3) {
            return 0;
        }
        if (day == n) {
            return 1;
        }

        if (memo[day][absences][lates] != -1) {
            return memo[day][absences][lates];
        }

        int ways = 0;

        // next day can be a present
        ways = dfs(n, day + 1, absences, 0, memo) % MOD;
        // or a absence if not already one
        ways = (ways + dfs(n, day + 1, absences + 1, 0, memo)) % MOD;
        
        // or a late
        ways = (ways + dfs(n, day + 1, absences, lates + 1, memo)) % MOD;
        
        
        memo[day][absences][lates] = ways;
        return ways;
    }

    int checkRecord(int n) {
        // given n days, how many possible valid attendance records

        // dp problem
        // for each day n, how many valid records
        // 3d dp with n, a, l

        int res = 0;

        vector<vector<vector<int>>> memo(n, vector<vector<int>>(2, vector<int>(3, -1)));
        // res = dfs(n, 0, 0, 0, memo);

        vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(3, vector<int>(4)));
        for (int d = 0; d <= n; ++d) {
            for (int a = 0; a <= 2; ++a) {
                for (int l = 0; l <= 3; ++l) {
                    if (a == 2 || l == 3) {
                        dp[d][a][l] = 0;
                    } else if (d == n) {
                        dp[d][a][l] = 1;
                    }
                }
            }
        }

        for (int d = n - 1; d >= 0; --d) {
            for (int a = 1; a >= 0; --a) {
                for (int l = 2; l >= 0; --l) {
                    int ways = 0;
                    ways = (ways + dp[d + 1][a][0]) % MOD;
                    ways = (ways + dp[d + 1][a + 1][0]) % MOD;
                    ways = (ways + dp[d + 1][a][l + 1]) % MOD;

                    dp[d][a][l] = ways;
                }
            }
        }

        res = dp[0][0][0];
        return res;
    }
};
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