<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 solve(string & s, int left, int right, int i){
        if(i == s.size()){
            if(left == right){
                return 1;
            }
            else return 0;
        }

        if(right > left) return 0;

        if(dp[left][right][i] != -1) return dp[left][right][i];

        if(s[i] == '('  ){
          return dp[left][right][i]= solve(s, left+1, right, i+1);
        }

        else if(s[i] == ')' ){
            return dp[left][right][i]= solve(s, left, right+1, i+1);
        }
        else {
            return dp[left][right][i]= (solve(s, left+1, right, i+1)|| solve(s, left, right+1, i+1)|| solve(s,left, right, i+1));

        }
    }
    bool checkValidString(string s) {
        memset(dp, -1, sizeof(dp));
        return solve(s, 0, 0, 0);
    }
};
11820