<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, s: str) -> int:
        # Initialize dp array to store the max length of substring ending with each character
        dp = [0] * 26
        
        # curLength will keep track of the length of the current valid substring
        curLength = 0
        
        for i in range(len(s)):
            # Calculate the index of the character in the dp array
            index = ord(s[i]) - ord('a')
            
            # If the current character follows the previous character in the alphabet (wraparound considered)
            if i > 0 and (ord(s[i]) - ord(s[i-1]) == 1 or (s[i-1] == 'z' and s[i] == 'a')):
                curLength += 1
            else:
                curLength = 1
            
            # Update the dp array for the current character
            dp[index] = max(dp[index], curLength)
        
        # The result is the sum of the dp array, which gives the count of unique substrings
        return sum(dp)
16900