<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 {
public:
    int findSubstringInWraproundString1(string p) {
        int n = p.size(), count=1;
        vector<int> arr(256,0);
        arr[p[0]] = 1;
        for(int i=1;i<n;i++)
        {
            if((p[i-1]=='z' && p[i]=='a') || p[i-1]+1 == p[i])
                count++;
            else
                count = 1;
            if(arr[p[i]] < count)
                arr[p[i]] = count;
        }
        return accumulate(arr.begin(),arr.end(),0);
    }
    int findSubstringInWraproundString(string s) {
        int maxLen[26] = {};

        int len = 0;
        int ci = 0;
        for(auto cc : s) {
            if(cc - 'a' == (ci + len) % 26)
                ++len;
            else {
                len = 1;
                ci = cc - 'a';
            }
            if(maxLen[cc - 'a'] < len)
                maxLen[cc - 'a'] = len;
        }
        
        int res = 0;
        for(int i = 0; i < 26; ++i)
            res += maxLen[i];
        return res;
    }
};
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