<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:
    unordered_map<string,int> dp;
    int helper(string ss){
        if(ss.length()==0)
            return 0;
        if(dp.find(ss)!=dp.end())
            return dp[ss];
        int cost = helper(ss.substr(0, ss.length()-1))+1;
        auto last_char = ss.back();
        for(int i=0; i<ss.length()-1; i++){
            if(ss[i]==last_char){
                cost = min(cost, helper(ss.substr(0, i+1))+helper(ss.substr(i+1, ss.length()-2-i)));
            }
        }
        dp[ss] = cost;
        return dp[ss];
    }
    int strangePrinter(string s) {
        return helper(s);
    }
};
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