minimum-remove-to-make-valid-parentheses.md

<p>Given a string <font face="monospace">s</font>&nbsp;of&nbsp;<code>&#39;(&#39;</code>&nbsp;,&nbsp;<code>&#39;)&#39;</code>&nbsp;and lowercase English characters.&nbsp;</p>

<p>Your task is to remove the minimum number of parentheses (&nbsp;<code>&#39;(&#39;</code>&nbsp;or&nbsp;<code>&#39;)&#39;</code>,&nbsp;in any positions ) so that the resulting <em>parentheses string</em> is valid and return <strong>any</strong> valid string.</p>

<p>Formally, a <em>parentheses string</em> is valid if and only if:</p>

<ul>
	<li>It is the empty string, contains only lowercase characters, or</li>
	<li>It can be written as&nbsp;<code>AB</code>&nbsp;(<code>A</code>&nbsp;concatenated with&nbsp;<code>B</code>), where&nbsp;<code>A</code>&nbsp;and&nbsp;<code>B</code>&nbsp;are valid strings, or</li>
	<li>It can be written as&nbsp;<code>(A)</code>, where&nbsp;<code>A</code>&nbsp;is a valid string.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;lee(t(c)o)de)&quot;
<strong>Output:</strong> &quot;lee(t(c)o)de&quot;
<strong>Explanation:</strong> &quot;lee(t(co)de)&quot; , &quot;lee(t(c)ode)&quot; would also be accepted.
</pre>

<p><strong>Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;a)b(c)d&quot;
<strong>Output:</strong> &quot;ab(c)d&quot;
</pre>

<p><strong>Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;))((&quot;
<strong>Output:</strong> &quot;&quot;
<strong>Explanation:</strong> An empty string is also valid.
</pre>

<p><strong>Example 4:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;(a(b(c)d)&quot;
<strong>Output:</strong> &quot;a(b(c)d)&quot;
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10^5</code></li>
	<li><code>s[i]</code>&nbsp;is one&nbsp;of&nbsp;&nbsp;<code>&#39;(&#39;</code> , <code>&#39;)&#39;</code> and&nbsp;lowercase English letters<code>.</code></li>
</ul>