<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:
    int removalCount(string s)
    {
        int ans = 0;
        int openCount = 0;
        for(int i = 0; i<s.size(); i++)
        {
            if(s[i] == '(')
                openCount++;
            else if(s[i] ==')')
            {
                if(openCount>0)
                    openCount--;
                else
                    ans++;
            }
        }
        ans += openCount;
        return ans;
    }
    void solve(string s, int k, unordered_map<string, int>&mp, vector<string>&ans)
    {
        if(mp.find(s)!=mp.end())
            return;
        mp[s]++;
        if(k == 0)
        {
            int count = removalCount(s);
            if(count == 0)
                ans.push_back(s);
        }

        for(int i = 0; i<s.size(); i++)
        {
            string first = s.substr(0, i);
            string second = s.substr(i+1);
            string temp = first+second;
            solve(temp, k-1, mp, ans);
        }
    }
    vector<string> removeInvalidParentheses(string s) {
        unordered_map<string, int>mp;
        int k = removalCount(s);
        vector<string>ans;
        solve(s, k, mp, ans);
        return ans;
    }
};
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