<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:
        dp = {}
        for i in range(len(s)):
            dp[(i, i)] = 1
            if i + 1 < len(s):
                dp[(i, i + 1)] = 1 if s[i] == s[i + 1] else 2
        def dfs(i, j):
            if (i, j) in dp:
                return dp[(i, j)]
            
            dp[(i, j)] = j - i + 1
            for dist in range(j - i):
                tmp = dfs(i, i + dist) + dfs(i + dist + 1, j)
                if s[i + dist] != s[j]:
                    dp[(i, j)] = min(dp[(i, j)], tmp)
                else:
                    dp[(i, j)] = min(dp[(i, j)], tmp - 1) 

            return dp[(i, j)]
        return dfs(0, len(s) - 1)
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