<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 checkAO(string input,int i,int j,int n,vector<vector<int>>&dp){
        if(i==n){
            return j==0;
        }
        if(j<0){
            return false;
        }
        if(dp[i][j]!=-1){
            return dp[i][j];
        }
        bool check=false;
        if(input[i]=='*'){
            check=(checkAO(input,i+1,j-1,n,dp) || checkAO(input,i+1,j+1,n,dp) || checkAO(input,i+1,j,n,dp));
        }
        else{
            if(input[i]=='('){
                check=checkAO(input,i+1,j+1,n,dp);
            }
            else{
                check=checkAO(input,i+1,j-1,n,dp);
            }
        }
        
        return dp[i][j]=check;
    }
    bool checkValidString(string s) {
        int n=s.size();
        vector<vector<int>>dp(n,vector<int>(2*n+1,-1));
        return checkAO(s,0,0,n,dp);
    }
};
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