Remove Invalid Parentheses
<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>
Breadth-First Search; String; Backtracking
class Solution {
public:
    vector<string> removeInvalidParentheses(string s) {
        unordered_set<string> ans;
        int l = 0, r = 0, n = s.size();
        for (char& c : s) {
            if (c == '(') {
                ++l;
            } else if (c == ')') {
                if (l) {
                    --l;
                } else {
                    ++r;
                }
            }
        }
        function<void(int, int, int, int, int, string)> dfs;
        dfs = [&](int i, int l, int r, int lcnt, int rcnt, string t) {
            if (i == n) {
                if (l == 0 && r == 0) {
                    ans.insert(t);
                }
                return;
            }
            if (n - i < l + r || lcnt < rcnt) {
                return;
            }
            if (s[i] == '(' && l) {
                dfs(i + 1, l - 1, r, lcnt, rcnt, t);
            }
            if (s[i] == ')' && r) {
                dfs(i + 1, l, r - 1, lcnt, rcnt, t);
            }
            int x = s[i] == '(' ? 1 : 0;
            int y = s[i] == ')' ? 1 : 0;
            dfs(i + 1, l, r, lcnt + x, rcnt + y, t + s[i]);
        };

        dfs(0, l, r, 0, 0, "");
        return vector<string>(ans.begin(), ans.end());
    }
};