<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 is_consec(self, c1,c2):
        if c1 == 'z' and c2 == 'a':
            return True
        if ord(c1) + 1 == ord(c2):
            return True
        return False

    def findSubstringInWraproundString(self, s: str) -> int:
        consec = 1
        out = {ord('a')+i : 0 for i in range(26)}
        for i in range(1, len(s)):
            curr, prev = s[i], s[i-1]
            if self.is_consec(prev, curr):
                consec += 1
            else:
                for j in range(i-consec, i):
                    out[ord(s[j])] = max(out[ord(s[j])], j - i + consec +1)
                consec = 1

        i = len(s)
        for j in range(i-consec, i):
            out[ord(s[j])] = max(out[ord(s[j])], j - i + consec +1)

        return sum([v for k, v in out.items()])



17100