<p>There is a strange printer with the following two special properties:</p>

<ul>
	<li>The printer can only print a sequence of <strong>the same character</strong> each time.</li>
	<li>At each turn, the printer can print new characters starting from and ending at any place and will cover the original existing characters.</li>
</ul>

<p>Given a string <code>s</code>, return <em>the minimum number of turns the printer needed to print it</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aaabbb&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Print &quot;aaa&quot; first and then print &quot;bbb&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Print &quot;aaa&quot; first and then print &quot;b&quot; from the second place of the string, which will cover the existing character &#39;a&#39;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

#define oo 1000000
#define MIN(_a, _b) (_a < _b ? _a:_b)

class Solution {
private:
	int dat[105][105];
	int n;
	
	void dfs(int s, int e, string str){
		if (s == e){
			dat[s][e] = 1;
			return;
		}
		
		if (dat[s][e] != -1){
			return;
		}
		
		if (str[s] == str[e]){
			if (dat[s][e-1] == -1)
				dfs(s, e - 1, str);
			dat[s][e] = dat[s][e - 1];
		}
		else{
			dat[s][e] = oo;
			for (int i = s; i <= e - 1; i++){
				if (dat[s][i] == -1)
					dfs(s, i, str);
				if (dat[i + 1][e] == -1)
					dfs(i + 1, e, str);
				dat[s][e] = MIN(dat[s][e], (dat[s][i] + dat[i + 1][e])); 
			}
		}
	}
	
public:
    int strangePrinter(string s) {
		n = s.length();
		
        for (int i = 0; i < n; i++){
            for (int j = 0; j < n; j++)
                dat[i][j] = -1;
        }
		dfs(0, n - 1, s);
		
		return dat[0][n-1];
    }
};
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