<p>We define the string <code>base</code> to be the infinite wraparound string of <code>&quot;abcdefghijklmnopqrstuvwxyz&quot;</code>, so <code>base</code> will look like this:</p>

<ul>
	<li><code>&quot;...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....&quot;</code>.</li>
</ul>

<p>Given a string <code>s</code>, return <em>the number of <strong>unique non-empty substrings</strong> of </em><code>s</code><em> are present in </em><code>base</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;a&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> Only the substring &quot;a&quot; of s is in base.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;cac&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are two substrings (&quot;a&quot;, &quot;c&quot;) of s in base.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;zab&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> There are six substrings (&quot;z&quot;, &quot;a&quot;, &quot;b&quot;, &quot;za&quot;, &quot;ab&quot;, and &quot;zab&quot;) of s in base.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def findSubstringInWraproundString(self, p: str) -> int:
        
        ia = list(map(lambda x:ord(x) - ord('a'), p))

        runs = {i: 0 for i in range(26)}
        st, r, la = ia[0], 1, ia[0]
        def proc():
            nonlocal st, r, la
            nonlocal runs
            for i in range(27):
                if runs[st]<r:
                    runs[st]=r
                st, r = (st+1)%26, r - 1
                if r<=0:
                    break


        for i in ia[1:]:
            if i == la + 1 or la == 25 and i == 0:
                st, r, la = st, r + 1, i
            else:
                proc()
                st, r, la = i, 1, i
        proc()

        ret = 0
        for i in runs.values():
            ret += i

        return ret


19000