<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 strangePrinter(self, s: str) -> int:
        
        #first remove the duplicates since they don't change anything in the number of turns 
        string = s[0]
        for i in range(1, len(s)): 
            if s[i] != string[-1]: 
                string += s[i] 
                
        dp = {} 

        def solve(start, end):
            if start > end: return 0 

            if (start, end) in dp: 
                return dp[(start, end)]
            
            ans = 1 + solve(start+1, end)
            for i in range(start+1, end+1): 
                if string[i] == string[start]: 
                    ans = min(ans, solve(start, i-1) + solve(i+1, end))

            dp[(start, end)] = ans 
            return ans 

        return solve(0, len(string)-1)


18800