<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:
        # Use memoization to store the results of previously computed states
        memo = {}
        return self.f(s, 0, 0, memo)
    
    def f(self, s, index, count, memo):
        n = len(s)
        
        # If count is negative, it's invalid (too many closing parentheses)
        if count < 0:
            return False
        
        # If we have reached the end of the string, check if the count is 0 (balanced)
        if index == n:
            return count == 0
        
        # Memoization key
        key = (index, count)
        
        # If the result for this state has already been computed, return it
        if key in memo:
            return memo[key]
        
        # If the current character is an opening parenthesis, increase count
        if s[index] == '(':
            result = self.f(s, index + 1, count + 1, memo)
        
        # If the current character is a closing parenthesis, decrease count
        elif s[index] == ')':
            result = self.f(s, index + 1, count - 1, memo)
        
        # If the current character is '*', it can represent '(', ')', or an empty string
        else:
            result = (self.f(s, index + 1, count + 1, memo) or  # Treat '*' as '('
                      self.f(s, index + 1, count - 1, memo) or  # Treat '*' as ')'
                      self.f(s, index + 1, count, memo))        # Treat '*' as empty
        
        # Store the result in the memo table
        memo[key] = result
        return result

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