<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>

#include <vector>
#include <string>
using namespace std;

class Solution {
public:
    int findSubstringInWraproundString(string s) {
        if (s.empty()) {
            return 0;
        }

        // Array to store the maximum length of substring ending with each letter
        vector<int> maxLength(26, 0);

        int currLength = 0;  // Length of current valid substring

        for (int i = 0; i < s.size(); i++) {
            // Check if the current character forms a wraparound sequence with the previous one
            if (i > 0 && (s[i] == s[i - 1] + 1 || (s[i - 1] == 'z' && s[i] == 'a'))) {
                currLength++;
            } else {
                currLength = 1;  // Reset length if not in a wraparound sequence
            }

            // Update the maximum length of substring ending with s[i]
            int index = s[i] - 'a';
            maxLength[index] = max(maxLength[index], currLength);
        }

        // Sum up all values in maxLength to get the total count of unique substrings
        int substr_Count = 0;
        for (int len : maxLength) {
            substr_Count += len;
        }

        return substr_Count;
    }
};
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