<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:
        maxans = 0
        dp = [0] * len(s)
        for i in range(1, len(s)):
            if s[i] == ")":
                if s[i - 1] == "(":
                    dp[i] = (dp[i - 2] if i >= 2 else 0) + 2
                elif i - dp[i - 1] > 0 and s[i - dp[i - 1] - 1] == "(":
                    dp[i] = (
                        dp[i - 1]
                        + (dp[i - dp[i - 1] - 2] if i - dp[i - 1] >= 2 else 0)
                        + 2
                    )
                maxans = max(maxans, dp[i])
        return maxans
16700