<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 {
public:
    map<pair<int,int>, int> m;
    int strangePrinter(string s) {
        string newS = RemoveDuplicate(s);
        return minTurns(newS, 0, newS.length() - 1);
    }

    int minTurns(string &s, int start, int end)
    {
        if (m.count({start,end}) != 0)
            return m[{start,end}];

        if (start > end)
            return 0;
        // Initialize with worst case: print each character separately
        int ans = 1 + minTurns(s, start+1, end);

        for(int i=start+1; i<=end; ++i)
        {
            if (s[i] != s[start])
                continue;
            int tmpAns = minTurns(s, start, i-1) + minTurns(s, i+1, end);
            ans = min(tmpAns, ans);
        }

        m.insert({{start, end}, ans});
        return ans;
    }

    string RemoveDuplicate(string &s)
    {
        string ans;
        int i=0;
        while(i < s.length())
        {
            char cur = s[i];
            ans += cur;
            while(i < s.length() && s[i] == cur)
                i += 1;
        }

        return ans;
    }
};
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