<p>Given the strings <code>s1</code> and <code>s2</code> of size <code>n</code> and the string <code>evil</code>, return <em>the number of <strong>good</strong> strings</em>.</p>

<p>A <strong>good</strong> string has size <code>n</code>, it is alphabetically greater than or equal to <code>s1</code>, it is alphabetically smaller than or equal to <code>s2</code>, and it does not contain the string <code>evil</code> as a substring. Since the answer can be a huge number, return this <strong>modulo</strong> <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, s1 = &quot;aa&quot;, s2 = &quot;da&quot;, evil = &quot;b&quot;
<strong>Output:</strong> 51 
<strong>Explanation:</strong> There are 25 good strings starting with &#39;a&#39;: &quot;aa&quot;,&quot;ac&quot;,&quot;ad&quot;,...,&quot;az&quot;. Then there are 25 good strings starting with &#39;c&#39;: &quot;ca&quot;,&quot;cc&quot;,&quot;cd&quot;,...,&quot;cz&quot; and finally there is one good string starting with &#39;d&#39;: &quot;da&quot;.&nbsp;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 8, s1 = &quot;leetcode&quot;, s2 = &quot;leetgoes&quot;, evil = &quot;leet&quot;
<strong>Output:</strong> 0 
<strong>Explanation:</strong> All strings greater than or equal to s1 and smaller than or equal to s2 start with the prefix &quot;leet&quot;, therefore, there is not any good string.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 2, s1 = &quot;gx&quot;, s2 = &quot;gz&quot;, evil = &quot;x&quot;
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>s1.length == n</code></li>
	<li><code>s2.length == n</code></li>
	<li><code>s1 &lt;= s2</code></li>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>1 &lt;= evil.length &lt;= 50</code></li>
	<li>All strings consist of lowercase English letters.</li>
</ul>

const int mod = 1e9+7;
class Solution {
    long long dp[501][51][2][2]={};
public:
    Solution(){
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
    }
    int findGoodStrings(int n, string s1, string s2, string evil) {
        return dfs(0, 0, true, true, n, s1, s2, evil, kmpArray(evil));
    }
private:
    int dfs(int pos, int evilMatched, bool boundLower, bool boundUpper, const int& n, const string& s1, const string& s2, const string& evil, const vector<int>& kmp){
        if(evilMatched == evil.size())
            return 0;
        if(pos == n)
            return 1;
        if(dp[pos][evilMatched][boundLower][boundUpper])
            return dp[pos][evilMatched][boundLower][boundUpper];
        
        char lo = boundLower ? s1[pos] : 'a';
        char hi = boundUpper ? s2[pos] : 'z';
        
        long long ans=0;
        for(char c=lo; c<=hi; ++c){
            int j = evilMatched;
            while(j && evil[j] != c)
                j = kmp[j-1];
            if(evil[j] == c)
                ++j;
            ans += dfs(pos+1, j, boundLower && (c == lo), boundUpper && (c == hi), n, s1, s2, evil, kmp);
            ans %= mod;
        }
        return dp[pos][evilMatched][boundLower][boundUpper] = ans;
    }
    vector<int> kmpArray(const string& evil){
        int i=1, len=0;
        const int n = evil.size();
        vector<int> kmp(n);
        
        while(i < n){
            if(evil[i] == evil[len])
                kmp[i++] = ++len;
            else{
                if(len)
                    len = kmp[len-1];
                else
                    ++i;
            }
        }
        return kmp;
    }
};
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