<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:
    //aabbaabbaa
    // string to min numbers of turns 
    unordered_map<string, int> memo;

    int helper(string s){
        if(memo.find(s) != memo.end()) return memo[s];
        if(s.empty()) return 0;
        if(s.size() == 1) return 1;
        
        /*bool uniq = true;
        for(int i = 1; i < s.size(); i++){
            if(s[i] != s[0]) uniq = false;
        }
        if(uniq) return memo[s] = 1;*/
        // if not repeated
        int ans = 1 + helper(s.substr(1));
        for(int i = 1; i < s.size(); i++){
            if(s[i] == s[0]){
                int res = helper(s.substr(0, i)) + helper(s.substr(i+1));
                ans = min(res, ans);
            }
        }
        return memo[s] = ans;
    }
    
    int strangePrinter(string s) {
        return helper(s);
    }
};
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