<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:
        # DP memoization table to store results for (idx, count)
        memo = {}

        def helper(char, count):
            if char == '(':
                count += 1
            elif char == ')':
                count -= 1
            return count

        def recFunction(s, idx, count):
            # Base case: if we've processed the whole string
            if idx == len(s):
                return count == 0  # It's valid if count of open '(' is balanced

            # If we have already computed this state, return the cached result
            if (idx, count) in memo:
                return memo[(idx, count)]

            # If count becomes negative, we have more ')' than '(', so it's invalid
            if count < 0:
                return False

            if s[idx] == '*':
                # Try all three possibilities for '*': as '(', ')', or ''
                case1 = helper('(', count)
                case2 = helper(')', count)
                if (recFunction(s, idx + 1, case1) or 
                    recFunction(s, idx + 1, case2) or 
                    recFunction(s, idx + 1, count)):
                    memo[(idx, count)] = True
                    return True
            elif s[idx] == '(':
                case1 = helper('(', count)
                if recFunction(s, idx + 1, case1):
                    memo[(idx, count)] = True
                    return True
            else:  # s[idx] == ')'
                case2 = helper(')', count)
                if recFunction(s, idx + 1, case2):
                    memo[(idx, count)] = True
                    return True

            # If none of the above options work, mark this state as False
            memo[(idx, count)] = False
            return False

        return recFunction(s, 0, 0)
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