<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:
    bool checkValidString(string s) {
        int n = s.length();
        
        // We only need two 1D arrays for the current and next row
        vector<bool> curr(n+1, false);
        vector<bool> next(n+1, false);
        
        // Base case: when we've processed the entire string and count == 0
        next[0] = true;
        
        // Fill from the end of the string to the beginning
        for(int i = n - 1; i >= 0; i--) {
            for(int count = 0; count <= n; count++) {
                if(s[i] == '(') {
                    if(count + 1 <= n) {
                        curr[count] = next[count+1];
                    } else {
                        curr[count] = false;
                    }
                } 
                else if(s[i] == ')') {
                    if(count > 0) {
                        curr[count] = next[count-1];
                    } else {
                        curr[count] = false;
                    }
                } 
                else {  // If it's '*', try all three possibilities: '(', ')', or empty
                    curr[count] = next[count];  // Treat '*' as empty
                    if(count + 1 <= n) curr[count] = curr[count] || next[count+1];  // Treat '*' as '('
                    if(count > 0) curr[count] = curr[count] || next[count-1];  // Treat '*' as ')'
                }
            }
            // Move current row to next row
            next = curr;
        }
        
        return next[0];  // The result is stored in next[0]
    }
};

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