<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:

        '''
        Wrong Solution: greedy divide and conquere O(nlogn)
        '''
        # cache = {}
        # def helper(seg):
        #     counters = Counter(seg)
        #     if len(counters) <= 1:
        #         return len(counters)
        #     if seg in cache:
        #         return cache[seg]

        #     rst = len(seg)
        #     for c in counters.keys():
        #         rst = min(rst, sum(helper(s) for s in seg.split(c))+1)
            
        #     cache[seg] = rst            
        #     return rst

        # rst = helper(s)
        # return rst


        '''
        Solution 2: Divide and conquere DP
        '''
        cache = {}
        sp = "".join([s[i] if i == 0 or s[i] != s[i-1] else "" for i in range(len(s))])
        def helper(i, j):
            if j < i:
                return 0
            if j == i:
                return 1
            if (i, j) in cache:
                return cache[(i, j)]
            rst = j-i+1
            for k in range(i, j):
                rst = min(rst, helper(i, k)+helper(k+1,j)-(1 if sp[i] == sp[j] else 0))  # note the -1 case!!!!!
            cache[(i, j)] = rst
            return rst

        return helper(0, len(sp)-1)
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