<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<string,int> lib;
    int strangePrinter(string s) {
        if(lib.find(s)!=lib.end())
            return lib[s];
        //cout<<s<<endl;
        int N = s.size();
        vector<int> dp(N+1);
        vector<vector<int>> pos(26, vector<int>());
        for(int i=0;i<N;i++){
            if(i>0 and s[i]==s[i-1]){
                dp[i+1] = dp[i];
            }else{
                dp[i+1] = dp[i]+1;
                for(auto& j:pos[s[i]-'a']){
                    dp[i+1] = min(dp[i+1], dp[j+1] + strangePrinter(s.substr(j+1, i-j-1)));
                }
            }

            pos[s[i]-'a'].push_back(i);  
        }

        lib[s]=dp.back();
        return dp.back();
    }
};
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