<p>An <strong>alphabetical continuous string</strong> is a string consisting of consecutive letters in the alphabet. In other words, it is any substring of the string <code>&quot;abcdefghijklmnopqrstuvwxyz&quot;</code>.</p>

<ul>
	<li>For example, <code>&quot;abc&quot;</code> is an alphabetical continuous string, while <code>&quot;acb&quot;</code> and <code>&quot;za&quot;</code> are not.</li>
</ul>

<p>Given a string <code>s</code> consisting of lowercase letters only, return the <em>length of the <strong>longest</strong> alphabetical continuous substring.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abacaba&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 4 distinct continuous substrings: &quot;a&quot;, &quot;b&quot;, &quot;c&quot; and &quot;ab&quot;.
&quot;ab&quot; is the longest continuous substring.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcde&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> &quot;abcde&quot; is the longest continuous substring.
</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 only English lowercase letters.</li>
</ul>

# 1st solution
# O(n) time | O(1) space
class Solution:
    def longestContinuousSubstring(self, s: str) -> int:
        start = 0
        prev = -1
        ans = 1
        for i, ch in enumerate(s):
            if prev < 0 or ord(ch) - 1 == prev:
                ans = max(ans, i - start + 1)
            else:
                start = i
            prev = ord(ch)
        return ans
158