<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:
    int f(int i, int sum, string s, vector<vector<int>>& dp) {
        if (i == 0) {
            if (sum == 1)
                return true;
            else
                return false;
        }
        if (sum == 0)
            return false;
        if (dp[i][sum] != -1)
            return dp[i][sum];
        if (s[i - 1] == '(')
            return dp[i][sum] = f(i - 1, sum - 1, s, dp);
        else if (s[i - 1] == ')')
            return dp[i][sum] = f(i - 1, sum + 1, s, dp);
        else if (s[i - 1] == '*') {
            bool a, b, c;
            a = f(i - 1, sum + 1, s, dp);
            b = f(i - 1, sum - 1, s, dp);
            c = f(i - 1, sum, s, dp);
            return dp[i][sum] = (a || b || c);
        }
        return dp[i][sum] = true;
    }
    bool checkValidString(string s) {
        int n = s.size();
        //vector<vector<int>> dp(n + 1, vector<int>(n + 2, 0));
        vector<int> prev(n + 2, 0);
        prev[1] = 1;
        for (int i = 1; i <= s.size(); i++) {
            vector<int> cur(n + 2, 0);
            for (int sum = s.size(); sum > 0; sum--) {
                if (s[i - 1] == '(')
                    cur[sum] = prev[sum - 1];
                else if (s[i - 1] == ')')
                    cur[sum] = prev[sum + 1];
                else {
                    bool a, b, c;
                    a = prev[sum + 1];
                    b = prev[sum - 1];
                    c = prev[sum];
                    cur[sum] = (a || b || c);
                }
            }
            prev=cur;
        }
        return prev[1];
    }
};
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