<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>

class Solution {
public:
    int findGoodStrings(int n, string s1, string s2, string evil) {
        const int MOD = 1e9 + 7;
        vector<int> lps = computeLPS(evil);
        vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(evil.size(), vector<int>(4, -1)));
        return dfs(0, 0, 1, 1, s1, s2, evil, lps, dp, MOD);
    }
    vector<int> computeLPS(const string& pattern) {
        int m = pattern.size();
        vector<int> lps(m, 0);
        for (int i = 1, len = 0; i < m;) {
            if (pattern[i] == pattern[len]) {
                lps[i++] = ++len;
            } else if (len) {
                len = lps[len - 1];
            } else {
                lps[i++] = 0;
            }
        }
        return lps;
    }
    int dfs(int pos, int evilMatched, int leftBound, int rightBound, const string& s1, const string& s2, const string& evil, const vector<int>& lps, vector<vector<vector<int>>>& dp, const int MOD) {
        if (evilMatched == evil.size()) return 0;
        if (pos == s1.size()) return 1;
        if (dp[pos][evilMatched][leftBound * 2 + rightBound] != -1) return dp[pos][evilMatched][leftBound * 2 + rightBound];
        char from = leftBound ? s1[pos] : 'a';
        char to = rightBound ? s2[pos] : 'z';
        int res = 0;
        for (char c = from; c <= to; ++c) {
            int j = evilMatched;
            while (j > 0 && c != evil[j]) j = lps[j - 1];
            if (c == evil[j]) ++j;
            res = (res + dfs(pos + 1, j, leftBound && (c == from), rightBound && (c == to), s1, s2, evil, lps, dp, MOD)) % MOD;
        }
        return dp[pos][evilMatched][leftBound * 2 + rightBound] = res;
    }
};
12500