<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>

class Solution:
    def printercalc(self, s, matrix, begin, end):
        if begin == end:
            matrix[begin][end] = 1
            return 1
        elif matrix[begin][end] != 0:
            return matrix[begin][end]
        
        matrix[begin][end] = len(s) * 2
        temp = -1
        
        for i in range(begin, end):
            if s[i] == s[end]:
                pass
            elif temp == -1:
                temp = i
            
            matrix[begin][end] = min(matrix[begin][end], self.printercalc(s, matrix, temp, i) + self.printercalc(s, matrix, i+1, end))

        if temp == -1:
            matrix[begin][end] = 1
            
        return matrix[begin][end]

                
    def strangePrinter(self, s: str) -> int:
        calc = []
        for i in range(len(s)):
            calc.append([])
            for j in range(len(s)):
                if i == j:
                    calc[i].append(1)
                else:
                    calc[i].append(0)
        return self.printercalc(s, calc, 0, len(s)-1)
16800