<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:
   // Memoization table to store results for (index, open) pairs
    unordered_map<int, unordered_map<int, bool>> memo;

    // Recursive function with memoization
    bool solve(const string &s, int i, int open) {
        // Base case: if we've reached the end of the string
        if (i == s.length()) {
            return open == 0;  // Valid if no unmatched open parentheses remain
        }

        // If already computed this state, return the stored result
        if (memo[i].count(open)) {
            return memo[i][open];
        }

        // If open parentheses become negative, it's invalid
        if (open < 0) {
            return false;
        }

        bool result = false;
        
        // Current character
        char ch = s[i];

        if (ch == '(') {
            // Treat '(' as an opening parenthesis: increment `open`
            result = solve(s, i + 1, open + 1);
        } else if (ch == ')') {
            // Treat ')' as a closing parenthesis: decrement `open`
            result = solve(s, i + 1, open - 1);
        } else if (ch == '*') {
            // Three possibilities for '*': treat it as '(', ')', or empty string
            result = solve(s, i + 1, open + 1)    // Treat '*' as '('
                  || solve(s, i + 1, open - 1)    // Treat '*' as ')'
                  || solve(s, i + 1, open);       // Treat '*' as an empty string
        }

        // Store the result in the memo table
        memo[i][open] = result;
        return result;
    }
    bool checkValidString(string s) {
        stack<char>sp;
       return solve(s,0,0); 
   
    } 
};
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