<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:
        def _bu(s):
            s_ = ""
            for c in s:
                if not s_ or s_[-1]!=c:
                    s_ += c

            n = len(s_)
            dp = [[1 if i == j or j - i == 1 and s_[i] == s_[j] else j - i + 1 for j in range(n)] 
                for i in range(n)]

            for l in range(2, n):
                for i in range(n - l):                
                    for k in range(i, i+l):
                        left = dp[i][k]
                        right = dp[k+1][i+l] if k < i + l else 0
                        cost = left + right - (s_[i] == s_[i+l])
                        dp[i][i + l] = min(dp[i][i+l], cost)
            
            return dp[0][-1]

        def _td(s):
            @cache 
            def _dfs(i, j):
                if i > j:
                    return  0
                if i == j:
                    return 1
                ans = j - i + 1
                for k in range(i, j):
                    ans = min(ans, _dfs(i, k) + _dfs(k+1, j) - (s[i] == s[j]))
                
                return ans
        
            s_ = ""
            for c in s:
                if not s_ or c != s_[-1]:
                    s_ += c
            s = s_

            return _dfs(0, len(s)-1)

        return _td(s)
20200