<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:
        s_len = len(s)

        # each cell [i, j] represents whether the string s[i:] is valid given previously j unclosed opening parens
        # our eventual goal is to find out if dp[0][0] aka s[0:] is valid given 0 previous unclosed opening parens (which is always the starting state of any string)
        dp = [[False] * (s_len + 1) for _ in range(s_len + 1)]
        dp[-1][0] = True

        for i in range(s_len - 1, -1, -1):
            for unclosed in range(s_len + 1):
                match s[i]:
                    case ")":  # reduce the number of unclosed by 1
                        if unclosed - 1 >= 0:
                            dp[i][unclosed] = dp[i + 1][unclosed - 1]
                    case "(":  # increase the number of unclosed by 1
                        if unclosed + 1 <= s_len:
                            dp[i][unclosed] = dp[i + 1][unclosed + 1]
                    case _:  # increase 1/ decrease 1/ maintain
                        if unclosed - 1 >= 0 and dp[i + 1][unclosed - 1]: # use ")"
                            dp[i][unclosed] = dp[i + 1][unclosed - 1]
                            continue
                        elif unclosed + 1 <= s_len and dp[i + 1][unclosed + 1]:   # use "("
                            dp[i][unclosed] = dp[i + 1][unclosed + 1]
                            continue
                        else:  # use ""
                            dp[i][unclosed] = dp[i + 1][unclosed]
        for line in dp:
            print(line)
        return dp[0][0]
16800