<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>

"""
Partition Logic:
base-case -> single char -> it takes 1 turn to print

- at each char divide the string into 2 parts

s[i] == s[j] -> that means that is sequence of chars 
-> reduce res by 1
"""
class Solution:
    def strangePrinter(self, s: str) -> int:
        cache = {}
        def dfs(i,j):
            #base_cases
            if i == j:
                return 1
            if (i,j) in cache:
                return cache[(i,j)]

            res = inf
            for k in range(i,j):
                res = min(res,dfs(i,k)+dfs(k+1,j))
            
            cache[(i,j)] = res-1 if s[i] == s[j] else res
            return cache[(i,j)]


        return dfs(0,len(s)-1)
20700