<p>Given a string containing just the characters <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>, return <em>the length of the longest valid (well-formed) parentheses </em><span data-keyword="substring-nonempty"><em>substring</em></span>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;(()&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The longest valid parentheses substring is &quot;()&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;)()())&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest valid parentheses substring is &quot;()()&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;&quot;
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= s.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>s[i]</code> is <code>&#39;(&#39;</code>, or <code>&#39;)&#39;</code>.</li>
</ul>


class Solution:
    def longestValidParentheses(self, s: str) -> int:

        if s.find('()')==-1 :
            return 0

        print('s0='+s)

        brackets = [-1]
        max_ = 0

        for i, el in enumerate(s):
            print('i0='+str(i))
            if el == '(':
                brackets.append(i)
            else:
                brackets.pop()
                if not brackets:
                    brackets.append(i)
                else:
                    #print('i1='+str(i))
                    #print('brackets[-1]='+str(brackets[-1]))
                    max_ = max(max_, i - brackets[-1])


        return max_

17100