<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>

class Solution:
    def minRemoveToMakeValid(self, s: str) -> str:
        
        count_open = 0
        count_close = 0

        ret_str = ""

        # Closing Parenthesis
        for i, c in enumerate(s):
            # letter
            if (c.isalpha()):
                ret_str += c

            # open case:
            elif c == "(":
                count_open += 1
                ret_str += c
            # close
            else:
                # matches 
                if count_open >= 1:
                    count_open -= 1
                    ret_str += c
                else:
                    # Invalid, not adding to result
                    continue

        # Opening Parenthesis
        ret_str_2 = ""
        for i in range(len(ret_str) - 1, -1, -1):
            c = ret_str[i] 
            # letter
            if (c.isalpha()):
                ret_str_2 += c
            # close case:
            elif c == ")":
                count_close += 1
                ret_str_2 += c
            # open
            else:
                # matches 
                if count_close >= 1:
                    count_close -= 1
                    ret_str_2 += c
                else:
                    # Invalid, not adding to result
                    continue

        return ret_str_2[::-1]



        # 2 pass solution  O(1)


        # stack O(n)

        # 

    #     class Solution:
    # def minRemoveToMakeValid(self, s: str) -> str:
    #     stack = []
    #     s = list(s)
    #     for i in range(len(s)):
    #         if s[i] == "(": stack.append(i)
    #         elif s[i] == ")":
    #             if stack: stack.pop()
    #             else: s[i] = ""
    #     for i in stack:
    #         s[i] = ""
    #     return "".join(s)
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