<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 checkRecord(int n) {
        // return checkRecord_TopDown(n);
        // return checkRecord_BottomUp(n);
        return checkRecord_BottomUp2(n);
    }

    int checkRecord_TopDown(int n) {
        // memo[len][absences][lates] := 
        //  #(valid strings we can further generate) of length len, given absences and lates now
        // -1 to indicate unsolved, NOT be 0 because it CAN NOT be distinguished from %MOD (which can be 0)
        vector<vector<vector<int>>> memo(n + 1, vector<vector<int>>(2, vector<int>(3, -1)));
        // for (int absences = 0; absences < 2; absences++)
        // for (int lates = 0; lates < 3; lates++)
        //     memo[0][absences][lates] = 1;
        return solve(n, 0, 0, memo);
    }

    int solve(int len, int absences, int lates, vector<vector<vector<int>>>& memo) {
        if (absences > 1 || lates > 2) return 0;
        if (len == 0) return 1;
        if (memo[len][absences][lates] != -1) {
            return memo[len][absences][lates];
        }
        int count = 0;
        // P
        count = (count + solve(len - 1, absences, 0, memo)) % MOD;
        // A
        count = (count + solve(len - 1, absences + 1, 0, memo)) % MOD;
        // L
        count = (count + solve(len - 1, absences, lates + 1, memo)) % MOD;
        return memo[len][absences][lates] = count;
    }

    int checkRecord_BottomUp(int n) {
        // dp[len][absences][lates] := 
        //  #(valid strings we could have generated) of length len containing absences and lates
        int kMOD = 1000000007;
        vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(2, vector<int>(3, 0)));
        dp[0][0][0] = 1;
        for (int len = 0; len < n; len++)
        for (int absences = 0; absences < 2; absences++)
        for (int lates = 0; lates < 3; lates++) {
            // P
            dp[len + 1][absences][0] = 
                (dp[len + 1][absences][0] + dp[len][absences][lates]) % MOD;
            // A
            if (absences < 1) {
                dp[len + 1][absences + 1][0] = 
                    (dp[len + 1][absences + 1][0] + dp[len][absences][lates]) % MOD;
            }
            // L
            if (lates < 2) {
                dp[len + 1][absences][lates + 1] = 
                    (dp[len + 1][absences][lates + 1] + dp[len][absences][lates]) % MOD;
            }
        }
        return accumulate(dp[n].begin(), dp[n].end(), 0, [&kMOD](long accu, auto& v) -> int {
            return (accu + accumulate(v.begin(), v.end(), 0L)) % kMOD;
        });
    }

    int checkRecord_BottomUp2(int n) {
        // dp[absences][lates] := 
        //  #(valid strings we could have generated) up to current length containing absences and lates
        int kMOD = 1000000007;
        vector<vector<int>> curr_dp(2, vector<int>(3, 0)), next_dp;
        curr_dp[0][0] = 1;
        for (int len = 0; len < n; len++) {
            // NOTE: outside of the loop of absences and lates!!!
            next_dp.assign(2, vector<int>(3, 0));

            for (int absences = 0; absences < 2; absences++)
            for (int lates = 0; lates < 3; lates++) {
                // P
                next_dp[absences][0] = 
                    (next_dp[absences][0] + curr_dp[absences][lates]) % kMOD;
                // A
                if (absences < 1)
                    next_dp[absences + 1][0] = 
                        (next_dp[absences + 1][0] + curr_dp[absences][lates]) % kMOD;
                // L
                if (lates < 2)
                    next_dp[absences][lates + 1] = 
                        (next_dp[absences][lates + 1] + curr_dp[absences][lates]) % kMOD;
            }

            curr_dp = next_dp;  // NOTE: outside of the loop of absences and lates!!!
        }
        // for (int absences = 0; absences < 2; absences++)
        // for (int lates = 0; lates < 3; lates++) {
        //     cout << next_dp[absences][lates] << endl;
        // }
        return accumulate(curr_dp.begin(), curr_dp.end(), 0, [&kMOD](long accu, auto& v) -> int {
            return (accu + accumulate(v.begin(), v.end(), 0L)) % kMOD;
        });
    }

private:
    int MOD = 1000000007;
};
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