<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:
    // Solution : 1 (Dynamic Programming)
    vector<vector<int> > dp;
    bool f(string s, int idx, int cnt){
        // 2 Base cases
        if (cnt < 0) return false;
        
        if (idx == s.size()) {
            return cnt == 0; 
        }

        // Memoization check
        if (dp[idx][cnt] != -1) return dp[idx][cnt];

        if (s[idx] == '(') {
            return dp[idx][cnt] = f(s, idx+1, cnt+1);
        }

        if (s[idx] == ')') {
            return dp[idx][cnt] = f(s, idx+1, cnt-1);
        }

        if (s[idx] == '*') {
            return dp[idx][cnt] = (f(s, idx+1, cnt+1) || f(s, idx+1, cnt-1) || f(s, idx+1, cnt+0));
        }

        return true;
    }

    // Solution : 2 (Greedy)

    bool greedy(string s){
        int mn = 0;
        int mx = 0;

        for(auto ch : s){
            if(ch == '('){
                mn += 1;
                mx += 1;
            } 
            else if(ch == ')'){
                mn -= 1;
                mx -= 1;
            }
            else {
                mn -= 1;
                mx += 1;
            }

            if(mn < 0) mn = 0;
            if(mx < 0) return false;
        }
        return mn == 0;
    }

    bool checkValidString(string s) {
        dp.clear();
        dp.resize(105,vector<int>(105, -1));
        // return f(s, 0, 0);
        return greedy(s);
    }
};
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