<p>Given a string <code>s</code> containing only three types of characters: <code>&#39;(&#39;</code>, <code>&#39;)&#39;</code> and <code>&#39;*&#39;</code>, return <code>true</code> <em>if</em> <code>s</code> <em>is <strong>valid</strong></em>.</p>

<p>The following rules define a <strong>valid</strong> string:</p>

<ul>
	<li>Any left parenthesis <code>&#39;(&#39;</code> must have a corresponding right parenthesis <code>&#39;)&#39;</code>.</li>
	<li>Any right parenthesis <code>&#39;)&#39;</code> must have a corresponding left parenthesis <code>&#39;(&#39;</code>.</li>
	<li>Left parenthesis <code>&#39;(&#39;</code> must go before the corresponding right parenthesis <code>&#39;)&#39;</code>.</li>
	<li><code>&#39;*&#39;</code> could be treated as a single right parenthesis <code>&#39;)&#39;</code> or a single left parenthesis <code>&#39;(&#39;</code> or an empty string <code>&quot;&quot;</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "()"
<strong>Output:</strong> true
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "(*)"
<strong>Output:</strong> true
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> s = "(*))"
<strong>Output:</strong> true
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s[i]</code> is <code>&#39;(&#39;</code>, <code>&#39;)&#39;</code> or <code>&#39;*&#39;</code>.</li>
</ul>

class Solution {
    std::unordered_map<std::string, bool> dp;
public:
    bool checkValidString(string s) {
        return check(s, 0, 0);
    }

    bool check(const std::string& s, int i, int count) {
        
        // key and check if we already have the state
        const std::string key = std::to_string(i) + "-" + std::to_string(count);
        if (dp.find(key) != dp.end()) return dp[key];

        // base
        if (count < 0) return false;
        if (i == s.size()) return count == 0;

        // normal advancement
        if (s[i] == '(') dp[key] = check(s, i+1, count+1);
        else if (s[i] == ')') dp[key] = check(s, i+1, count-1);
        // try all possibilities for *
        else {
            dp[key] = check(s, i+1, count+1) || check(s, i+1, count-1) || check(s, i+1, count);
        }

        // done, return state
        return dp[key];
    }
};
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