<p>Given a string <code>s</code> that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.</p>

<p>Return <em>a list of <strong>unique strings</strong> that are valid with the minimum number of removals</em>. You may return the answer in <strong>any order</strong>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;()())()&quot;
<strong>Output:</strong> [&quot;(())()&quot;,&quot;()()()&quot;]
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;(a)())()&quot;
<strong>Output:</strong> [&quot;(a())()&quot;,&quot;(a)()()&quot;]
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;)(&quot;
<strong>Output:</strong> [&quot;&quot;]
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 25</code></li>
	<li><code>s</code> consists of lowercase English letters and parentheses <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>.</li>
	<li>There will be at most <code>20</code> parentheses in <code>s</code>.</li>
</ul>

class Solution {
public:
    /* get relative count of '(' and ')' */
    int getParValue(string s)
    {
        int _count = 0;
        for (int i=0; i<s.size(); i++) {
            if (s[i] == '(') _count++;
            else if (s[i] == ')' && _count-- == 0) return _count;
        }
        return _count;
    }

    /**
    * Idea 1. BFS
    * remove min number of parenthese
    */
    vector<string> removeInvalidParentheses(string s)
    {
        vector<string> ans;
        queue<string> q;
        q.push(s);

        bool found = false;
        while (!q.empty()) {
            int len = q.size();
            while (len--) {
                string _top = q.front(); 
                q.pop();

                /* check validity */
                int par_val = getParValue(_top);
                char remove_char;
                if (par_val > 0) remove_char = '(';
                else if (par_val < 0) remove_char = ')';
                /* found valid str in this level */
                else {
                    found = true;
                    /* linear search for uniqueness */
                    bool existed = false;
                    for (int i=0; i<ans.size(); i++)
                        if (ans[i] == _top) {
                            existed = true;
                            break;
                        }
                    
                    if (!existed)
                        ans.push_back(_top);
                }

                if (found) continue;

                /* erase `remove_char` to form substr */
                for (int i=0; i<_top.size(); i++) {
                    /* uninqueness (not complete) */
                    if (_top[i] != remove_char || (i && _top[i-1] == _top[i])) continue;

                    q.push(
                        _top.substr(0, i) + _top.substr(i+1)
                    );
                }
            }
        }

        return ans;
    }

};
125015