<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(string diff,int s,int e,vector<vector<int>> &v)
{
    if(s==e)
    return 1;
    if(s>e)
    return 0;
    if(v[s][e]!=-1)
    return v[s][e];
    int a=1+solve(diff,s+1,e,v);
    for(int i=s+1;i<=e;i++)
    {
        int b=INT_MAX;
        if(diff[s]==diff[i])
        b=solve(diff,s,i-1,v)+solve(diff,i+1,e,v);
        a=min(a,b);

    }
    return v[s][e]=a;
}
    int strangePrinter(string s) {
        string diff="";
        diff+=s[0];
        for(int i=1;i<s.size();i++)
        {
            if(s[i]!=s[i-1])
            diff+=s[i];
            
        }
        int n=diff.size();
        vector<vector<int>> v(n,vector<int> (n,-1));

        return solve(diff,0,diff.size()-1,v);
    }
};
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