<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 strangePrinter(string s) {
        int dp[101][101][128];
        memset(dp,-1,sizeof dp);
        return solve(dp,s,0,s.size()-1,'z'+1);
    }
    
    int solve(int dp[101][101][128],string &s, int l, int r,char prev){
        if(l>r) return 0;
        if(dp[l][r][prev]!=-1) return dp[l][r][prev];
        //cout<<l<<" "<<r<<" "<<prev<<endl;
        int mini[27];
        
        int res=100000;
        memset(mini,-1,sizeof mini);
        for(int i=l;i<=r;i++){
            if(mini[s[i]-97]==-1){
                mini[s[i]-97]=i;
                if(prev!=s[i]) res=min(res,1+
                    solve(dp,s,l,mini[s[i]-97]-1,prev)+
                    solve(dp,s,i+1,r,prev)
                );
            }
            else if(prev!=s[i]){
                res=min(res,1+
                    solve(dp,s,l,mini[s[i]-97]-1,prev)+
                    solve(dp,s,mini[s[i]-97]+1,i-1,s[i])+
                    solve(dp,s,i+1,r,prev)
                );
            }
        }
        
        if(res==100000) res=0;
        
        return dp[l][r][prev]=res;
    }
};
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