<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:
    int solve(int start , int end, string &s, map<string,int> &mp){
        if(start > end) return 0;
        string key = to_string(start) + " " + to_string(end) ;
        if(mp.find(key) != mp.end()) return mp[key] ;
        
        int ans = 1 + solve(start+1,end,s,mp) ;
        for(int i = start+1 ; i <= end ; i++){
            if(s[i] == s[start]){
                int curr = solve(start+1,i-1,s,mp) + solve(i,end,s,mp) ;
                ans = min(curr , ans) ;
            } 
        }
        return mp[key] = ans ;
    }
    int strangePrinter(string s) {
        int start = 0 , end = s.size()-1;
        map<string ,int > mp ;
        return solve(start , end , s , mp) ;
    }
};
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