<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:

    bool isValid(string s) {
        int balance = 0;
        for (char c : s) {
            if (c == '(') balance++;
            else if (c == ')') {
                balance--;
                if (balance < 0) return false; // Early exit on invalidity
            }
        }
        return balance == 0;
    }

    unordered_map<string,int>mp;
    map<pair<string,int>,bool>dp;
    int maxi=0;
    void solve(string s,int i){
        if(maxi>s.size())return;
        if(i>=s.size()){
            if(isValid(s)){
                int sz = s.size();
                maxi=max(maxi,sz);
                mp[s]=sz;
            }
            return;
        }

        if (dp.find({s,i}) != dp.end()) {
            return;
        }

        solve(s,i+1);
        char ch = s[i];
        s.erase(s.begin()+i);
        solve(s,i);
        s.insert(s.begin()+i,ch);

        dp[{s,i}]=true;
    }

    vector<string> removeInvalidParentheses(string s) {
        solve(s,0);
        vector<string>ans;
        for(auto i:mp)maxi = max(maxi,i.second);
        for(auto i:mp){
            if(i.second==maxi)ans.push_back(i.first);
        }
        return ans;
    }
};
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