<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:
        # Remove consecutive duplicates as they are of no use
        new_string = [s[0]]
        for i in range(len(s)-1):
            if s[i] == s[i+1]:
                continue
            new_string.append(s[i+1])
        
        n = len(new_string)
        @cache
        def min_turns(start, end):
            if start > end:
                return 0
            if start == end:
                return 1
            
            ans = float('inf')
            for i in range(start, end):
                # abcd | efgh, abc | defg, ab | defgh, .....
                # cut the string at each point. (MCM)
                ans = min(ans, min_turns(start, i) + min_turns(i+1, end))
            
            if new_string[start] == new_string[end]:
                ans -= 1

            return ans

        return min_turns(0, n-1)
        
20300