<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:
        # Preprocess the string to remove consecutive duplicate characters
        s = "".join(char for char, _ in itertools.groupby(s))
        print(s)
        memo = {}

        def _minimum_turns(start, end) -> int:
            # Base case: empty string requires 0 turns
            if start > end:
                return 0

            # If result is memoized, return it
            if (start, end) in memo:
                return memo[(start, end)]

            # Initialize with worst case: print each character separately
            min_turns = 1 + _minimum_turns(start + 1, end)

            # Try to optimize by finding matching characters
            for k in range(start + 1, end + 1):
                if s[k] == s[start]:
                    # If match found, try splitting the problem
                    turns_with_match = _minimum_turns(
                        start, k - 1
                    ) + _minimum_turns(k + 1, end)
                    min_turns = min(min_turns, turns_with_match)

            # Memoize and return the result
            memo[(start, end)] = min_turns
            return min_turns

        # Start the recursive process
        return _minimum_turns(0, len(s) - 1)

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