<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:
    
    int dp[101][101][101];
    bool f(int idx,string &s,int n,int open,int close){

    if(idx==n){
        return open==close;
    }
    if(dp[idx][open][close]!=-1){
        return dp[idx][open][close];
    }

    if(s[idx]=='('){
       
       return dp[idx][open][close]=f(idx+1,s,n,open+1,close);
    }
    else if(s[idx]==')'){
       if(open<close+1){
        return false;
       }
       return dp[idx][open][close]=f(idx+1,s,n,open,close+1);
    }
    else{
       bool op1=false;
       if(open>close){
        op1=f(idx+1,s,n,open,close+1);
       }
       return dp[idx][open][close]=op1||f(idx+1,s,n,open+1,close)||f(idx+1,s,n,open,close);
    }

    }

    bool checkValidString(string s) {
        int n=s.length();
        memset(dp,-1,sizeof(dp));
        return f(0,s,n,0,0);

    }
};
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