<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:
    def checkValidString(self, s: str) -> bool:
        @cache
        def dfs(ind: int, opcnt: int) -> bool:
            """
            @param ind: current index of char being checked
            @param opcnt: number(count) of open parenthesis
            """
            if ind == len(s):
                # reach the end of string, need to check if number of open parenthesis is 0 
                return opcnt == 0
            
            if s[ind] == '(':
                # if current char being checked is open paren(i.e."("), we increase open paren count
                return dfs(ind + 1, opcnt + 1)
            elif s[ind] == ')':
                # if current char being checked is close paren(i.e.")"), open paren count should be greater than 0
                # otherwise, the current close paren has no match.
                return opcnt and dfs(ind + 1, opcnt - 1)
            else:
                # current char being checked is "*"; so there are 3 possibilities: open paren, close paren and empty
                return dfs(ind + 1, opcnt) or dfs(ind + 1, opcnt + 1) or (opcnt and dfs(ind + 1, opcnt - 1))
        
        return dfs(0, 0)


        
17200