<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:
        s = ''.join(a for a, b in zip(s, '#' + s) if a != b)
        n = len(s)
        dp = [[-1] * n for _ in range(n)]

        return self.Util(0, n - 1, s, dp)

    def Util(self, i: int, j: int, s: str, dp: list) -> int:
        if i > j:
            return 0

        if dp[i][j] != -1:
            return dp[i][j]

        first_letter = s[i]
        # If the current character is not repeated in the rest of the string
        answer = 1 + self.Util(i + 1, j, s, dp)
        for k in range(i + 1, j + 1):
            # If repeated then update the answer
            if s[k] == first_letter:       
                better_answer = self.Util(i, k - 1, s, dp) + self.Util(k + 1, j, s, dp)
                answer = min(answer, better_answer)
        dp[i][j] = answer
        return answer
16500