<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:
    // https://leetcode.com/problems/valid-parenthesis-string/submissions/1383871878/?envType=company&envId=facebook&favoriteSlug=facebook-all&status=TO_DO&difficulty=MEDIUM
    map<pair<int, int>, bool> memo;
    bool dfs(int idx, int cnt, string &s) {
        if(idx == (int)s.size()) {
            return cnt == 0;
        }

        if(cnt < 0) return false;

        if(memo.count({idx, cnt})) return memo[{idx, cnt}];
        
        bool ret = false;
        if(s[idx] == '(') {
            ret = dfs(idx+1, cnt+1, s);
        } else if(s[idx] == ')') {
            ret = dfs(idx+1, cnt-1, s);
        } else {
            ret = dfs(idx+1, cnt-1, s) || dfs(idx+1, cnt+1, s) || dfs(idx+1, cnt, s);
        }

        memo[{idx, cnt}] = ret;
        return ret;
    }

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