<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:
    unordered_map<string, bool> memo;

    bool solve(string &s, int i, int openCount) {
        if(openCount <0) return false;
        if (i == s.size()) {
            return openCount == 0;
        }

        string key = to_string(i) + "," + to_string(openCount);
        if (memo.find(key) != memo.end()) {
            return memo[key];
        }

        bool a = false , b = false , c = false;

        if (s[i] == '(') {
            a =  solve(s, i + 1, openCount + 1);
        }
        else if (s[i] == ')') {
             b = solve(s, i + 1, openCount - 1);
        }
        else if (s[i] == '*') {

            a = solve(s, i + 1, openCount + 1);
            b = solve(s, i + 1, openCount - 1);
            c = solve(s, i + 1, openCount);
        }

        memo[key] = a || b || c;
        return a || b || c ;
    }

    bool checkValidString(string s) {
        return solve(s, 0, 0);
    }
};

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