<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:
        
        # What is the minimal number of operations needed to print characters from start to end?
        # [start, end - 1] index inclusive
        @cache
        def dp(start, end):
            if start >= end:
                return 0
            
            # Consider the naive case of printing every character from start to end with 1 operation each
            cur_min = 1 + dp(start + 1, end)
            
            # Iterate through current string and consider cases where we can minimize operations by printing characters in one go
            for i in range(start + 1, end):
                # If the start and end characters are the same we can consider printing them in one go
                if s[start] == s[i]:
                    cur_min = min(
                        cur_min,
                        # Why?????
                        dp(start, i) + dp(i + 1, end)
                    )
            print(f'dp({start}, {end}): {cur_min}')
            return cur_min
        
        return dp(0, len(s))

19400