Strange Printer
<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>
String; Dynamic Programming
class Solution {
public:
    int strangePrinter(string s) {
        int n = s.size();
        int f[n][n];
        memset(f, 0x3f, sizeof(f));
        for (int i = n - 1; ~i; --i) {
            f[i][i] = 1;
            for (int j = i + 1; j < n; ++j) {
                if (s[i] == s[j]) {
                    f[i][j] = f[i][j - 1];
                } else {
                    for (int k = i; k < j; ++k) {
                        f[i][j] = min(f[i][j], f[i][k] + f[k + 1][j]);
                    }
                }
            }
        }
        return f[0][n - 1];
    }
};