<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:
string rem(string s){
    string temp="";
    int n = s.length();
    for(int i=1;i<s.length();i++){
        if(s[i]!=s[i-1]){
            temp+=s[i-1];
        }
    }
    temp+=s[n-1];
    return temp;
}
int f(int st,int end,string s,vector<vector<int>>&dp){
    if(st>end)return 0;
    if(dp[st][end]!=-1)return dp[st][end];
    int mini = 1+f(st+1,end,s,dp);
    for(int k=st+1;k<=end;k++){
        if(s[k]==s[st]){
            mini = min({mini,f(st,k-1,s,dp)+f(k+1,end,s,dp)});
        }
    }
    return dp[st][end] = mini;
}
    int strangePrinter(string s) {
      //abcaacbacb
      s = rem(s);
      int n = s.length();
      vector<vector<int>>dp(n+1,vector<int>(n+1,-1));
      return f(0,s.length()-1,s,dp);
    }
};
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