<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:
        # dp[i]: i is ord(char), i in s as the last character
        # how long to the left is the substring ending in i in s is also a substring in base
        # add all in dp to reach number of unique substrings in base string
        # step 1: calculate ord of each char in s
        sList = [ord(c) - ord('a') for c in s]
        dp = [0]*26
        dp[sList[0]] = 1
        cur_len = 1
        for i in range(1, len(sList)):
            cur = sList[i]
            pre = sList[i - 1]
            if (cur == 0 and pre == 25) or (cur == pre + 1):
                cur_len += 1
            else:
                cur_len = 1
            dp[cur] = max(dp[cur], cur_len)
        res = 0
        for n in dp:
            res += n
        return res

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