<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 dp[101][101][101];

    int f(int i,int j,int len,string &s){
        if(i>j) return 0;
        if(i==j) return 1;
        if(dp[i][j][len]!=-1) return dp[i][j][len];
        int x = 1 + f(i+1,j,0,s);
        for(int ind=i+1;ind<=j;ind++){
            if(s[ind]==s[i]){
                x=min(x,f(i+1,ind,0,s)+f(ind+1,j,len+1,s));
            }
        }
        return dp[i][j][len]=x;
    }
    int strangePrinter(string s) {
        memset(dp,-1,sizeof(dp));
        return f(0,s.length()-1,0,s);
    }
};
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