<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:
    string uniqueString(string& s) {
        string res = "";
        int i = 0;
        while (i < s.length()) {
            char cur = s[i];
            res += s[i];
            while (i < s.length() && s[i] == cur) {
                i++;
            }
        }
        return res;
    }
    int util(string s, unordered_map<string,int>& mp){
        if(s.length()==0) return 0;
        if(mp.find(s)!=mp.end()) return mp[s];
        int m  = 1+util(s.substr(1),mp);
        for(int i=1; i<s.length(); i++){
            if(s[i]==s[0])
                m = min(m, util(s.substr(0,i), mp)+util(s.substr(i+1), mp));
        }
        cout<<s<<m<<endl;
        mp[s] = m;
        return m;
    }
    int strangePrinter(string s) {
        unordered_map<string,int> mp;
        return util(uniqueString(s), mp);
    }
};
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