<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 {
public:
    bool checkValidString(string s) {
        // O(3^n) time, O(n) space
        std::vector<std::vector<std::pair<bool,bool>>> dp (s.size(), std::vector<std::pair<bool,bool>>(s.size(), {false, false})); // <open_bracket, true or false>
       return exploreValidString(s, dp, 0, 0);
    }

private:
    [[nodiscard]] bool exploreValidString(const std::string& s, std::vector<std::vector<std::pair<bool,bool>>>& dp, int i, int open_bracket)
    {
        if(open_bracket < 0) return false;

        for(; i < s.size(); i++)
        {
            if(s[i] == '(')
            {
                open_bracket++;
            }
            else if (s[i] == ')')
            {
                open_bracket--;
            }
            else if (s[i] == '*')
            {
                // consider it as an open bracket, closed bracket or an empty space, see if we have already checked at this combination
                if(dp[i][open_bracket].first) 
                {
                    return dp[i][open_bracket].second;
                }
                else
                {
                    dp[i][open_bracket] = {true, exploreValidString(s, dp, i+1, open_bracket+1) || exploreValidString(s, dp, i+1, open_bracket-1) || exploreValidString(s, dp, i+1, open_bracket)};
                    return dp[i][open_bracket].second;
                }
            }

            if(open_bracket < 0) return false;
        }

        return open_bracket == 0;
    }
};
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