<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>

class Solution: # my solution
    def longestContinuousSubstring(self, s: str) -> int:
        if len(s) == 1:
            return 1
        i = 1
        longest = 1
        prev, cur_l = s[0], 1
        while i < len(s):
            if (ord(s[i]) - ord(prev)) == 1:
                cur_l += 1
                longest = max(longest, cur_l)
            else:
                cur_l = 1
            prev = s[i]
            i+=1

        return longest




# class Solution:  # 2 pointers
#     def longestContinuousSubstring(self, s: str) -> int:
#         i = 0
#         j = 1
#         n = len(s)
#         ans = 1
#         while j<n:
#             while j<n and ord(s[j])-ord(s[j-1])==1:
#                 ans = max(ans,j-i+1)
#                 j+=1
#             if j<n and ord(s[j])-ord(s[j-1])!=1:
#                 i = j
#                 j+=1

#         return ans
203