<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 f(int n, string &s, int check, vector<vector<int>> &dp){
        if(n<0){
            if(check == 0)
                return true;   
            return false;
        }
        if(check < 0) return false;
        if(dp[n][check] != -1) return dp[n][check];
        bool ans;
        if(s[n] == ')')
            ans = f(n-1, s, check + 1, dp);
        else if(s[n] == '(')
            ans = f(n-1, s, check-1, dp);
        else{
            bool left = f(n-1, s, check - 1, dp);
            bool right = f(n-1, s, check + 1, dp);
            bool mid = f(n-1, s, check, dp);

            ans = left | right | mid;
        }
        dp[n][check] = ans;
        return ans;
    }
    bool checkValidString(string s) {
        int n = s.size();
        vector<vector<int>> dp(n, vector<int> (101, -1));
        return f(n-1, s, 0, dp);
    }
};
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