<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:
        n = len(s)
        memo = [[0] * n for _ in range(n)]
        prev = [[0] * n for _ in range(26)]

        for i in range(n):
            memo[i][i] = 1
        
        prev[ord(s[0]) - ord('a')][0] = 1

        def calc_range(start, end):
            if memo[start][end] > 0:
                return memo[start][end]
            
            for i in range(start + 1, end + 1):
                char = s[i]
                ind = ord(char) - ord('a')
                if char == s[i - 1]:
                    memo[start][i] = memo[start][i - 1]
                else:
                    memo[start][i] = memo[start][i - 1] + 1
                    for j in range(start, i):
                        if prev[ind][j]:
                            memo[start][i] = min(memo[start][i], memo[start][j] + calc_range(j + 1, i - 1))

                prev[ind][i] = 1
        
            return memo[start][end]
        
        
        ans = calc_range(0, n - 1)
        return ans
17900