Minimum Remove to Make Valid Parentheses
<p>Given a string <font face="monospace">s</font> of <code>&#39;(&#39;</code> , <code>&#39;)&#39;</code> and lowercase English characters.</p>

<p>Your task is to remove the minimum number of parentheses ( <code>&#39;(&#39;</code> or <code>&#39;)&#39;</code>, 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 <code>AB</code> (<code>A</code> concatenated with <code>B</code>), where <code>A</code> and <code>B</code> are valid strings, or</li>
	<li>It can be written as <code>(A)</code>, where <code>A</code> is a valid string.</li>
</ul>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either&nbsp;<code>&#39;(&#39;</code> , <code>&#39;)&#39;</code>, or lowercase English letter.</li>
</ul>
Stack; String
class Solution:
    def minRemoveToMakeValid(self, s: str) -> str:
        stk = []
        x = 0
        for c in s:
            if c == ')' and x == 0:
                continue
            if c == '(':
                x += 1
            elif c == ')':
                x -= 1
            stk.append(c)
        x = 0
        ans = []
        for c in stk[::-1]:
            if c == '(' and x == 0:
                continue
            if c == ')':
                x += 1
            elif c == '(':
                x -= 1
            ans.append(c)
        return ''.join(ans[::-1])
