<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, s1, s2, evil):
        M = 10 ** 9 + 7
        m = len(evil)
        memo = {}
        # use kmp to fetch the length of longest common prefix
        dfa = [0] * m
        l = 0
        for r in range(1, m):
            while l and evil[l] != evil[r]:
                l = dfa[l - 1]
            l += evil[l] == evil[r]
            dfa[r] = l

        def dfs(i, x, bound):
            if x == m:
                return 0
            if i == n:
                return 1
            if (i, x, bound) not in memo:
                cnt = 0
                lo = ord('a' if bound & 1 else s1[i])  # "be*" -> {"be*", "ca*", ..}, 'b' when bound bit = ?0
                hi = ord('z' if bound & 2 else s2[i])  # "do*" -> {"cz*", "do*", ..}, 'd' when bound bit = 0?
                for j, c in enumerate(chr(o) for o in range(lo, hi + 1)):
                    y = x
                    while y and evil[y] != c:
                        y = dfa[y - 1]
                    y += evil[y] == c
                    cnt = (cnt + dfs(i + 1, y, bound | (j > 0) | (j < hi - lo) << 1)) % M
                memo[i, x, bound] = cnt
            return memo[i, x, bound]
			
        return dfs(0, 0, 0)

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