<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>

#pragma gcc optimize("03")
auto init = [](){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 'c';
}();
class Solution {
    const int mod = 1e9+7;
    int solve(int n , int i  ,int absent , int late , vector<vector<vector<int>>>&dp){
        if(i==n)return 1;
        int ans =0;
        if(dp[i][absent][late]!=-1)return dp[i][absent][late];
        ans += solve(n , i+1 , absent , 0 , dp);
        if(absent<1) ans = (ans + solve(n , i+1 , absent+1 , 0,dp))%mod;
        if(late<2)ans = (ans + solve(n , i+1 , absent , late+1,dp))%mod;
        return dp[i][absent][late] = ans;
    }
    int solvetab(int n){
        vector<vector<vector<int>>>dp(n+1, vector<vector<int>>(2 , vector<int>(3,0)));
        for(int i = 0 ; i<2 ; i++){
            for(int j=0 ; j<3 ; j++){
                dp[n][i][j] = 1;
            }
        }
        
        for(int i = n-1 ; i>=0 ; i--){
            for(int absent = 1 ; absent>=0 ; absent--){
                for(int late = 2 ; late>=0 ; late--){
                    int ans =0;
                    ans =  (ans + dp [i+1] [absent][0 ])%mod;
                    if(absent<1) ans = (ans + dp[ i+1 ][ absent+1] [0])%mod;
                    if(late<2)ans = (ans + dp[i+1][ absent][late+1])%mod;
                    dp[i][absent][late] = ans;
        }
        }
        }
        return dp[0][0][0];
    }
      int solvesp(int n){
        vector<vector<int>>curr(2 , vector<int>(3,0));
        vector<vector<int>>next(2 , vector<int>(3,0));
        for(int i = 0 ; i<2 ; i++){
            for(int j=0 ; j<3 ; j++){
                next[i][j] = 1;
            }
        }
        
        for(int i = n-1 ; i>=0 ; i--){
            for(int absent = 1 ; absent>=0 ; absent--){
                for(int late = 2 ; late>=0 ; late--){
                    int ans =0;
                    ans =  (ans + next[absent][0 ])%mod;
                    if(absent<1) ans = (ans + next[ absent+1] [0])%mod;
                    if(late<2)ans = (ans + next[ absent][late+1])%mod;
                    curr[absent][late] = ans;
        }
        }
        next = curr;
        }
        return curr[0][0];
    }
public:
    int checkRecord(int n) {
        vector<vector<vector<int>>>dp(n+1, vector<vector<int>>(2 , vector<int>(3,-1)));
        // return solve(n , 0  , 0 , 0  , dp);
        return solvesp(n);
    }
};




313871