<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:
    def findGoodStrings(self, n: int, s1: str, s2: str, evil: str) -> int:
        MOD = 10**9 + 7
        m = len(evil)

        def buildKMP(pattern):
            kmp = [0] * m
            j = 0
            for i in range(1, m):
                while j > 0 and pattern[i] != pattern[j]:
                    j = kmp[j - 1]
                if pattern[i] == pattern[j]:
                    j += 1
                    kmp[i] = j
            return kmp

        kmp = buildKMP(evil)

        @lru_cache(None)
        def dp(i, tight1, tight2, evilLen):
            if evilLen == m:
                return 0
            if i == n:
                return 1

            lo = s1[i] if tight1 else 'a'
            hi = s2[i] if tight2 else 'z'

            res = 0
            for c in range(ord(lo), ord(hi) + 1):
                ch = chr(c)
                # Update the length of the evil prefix
                nextEvilLen = evilLen
                while nextEvilLen > 0 and evil[nextEvilLen] != ch:
                    nextEvilLen = kmp[nextEvilLen - 1]
                if evil[nextEvilLen] == ch:
                    nextEvilLen += 1
                res = (res + dp(i + 1, tight1 and ch == lo, tight2 and ch == hi, nextEvilLen)) % MOD
            return res

        return dp(0, True, True, 0)
22000