<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 n=s.size();
        vector<vector<int>> dp(n,vector<int>(n,1e9));
        for(int i=0;i<n;i++){
            dp[i][i]=1;
        }
        for(int j=1;j<n;j++){
            for(int i=0;i+j<n;i++){
                for(int k=i;k<i+j;k++){
                    int x=0;
                    if(dp[i][k]==1 && dp[k+1][i+j]==1 && s[i]==s[k+1]) dp[i][i+j]=min(dp[i][i+j],1);
                    else if(dp[i][k]==1 && dp[k+1][i+j]==1) dp[i][i+j]=min(dp[i][i+j],2);
                    else if(dp[i][k]==1 && s[k]==s[k+1]) dp[i][i+j]=min(dp[i][i+j],dp[k+1][i+j]);
                    else if(dp[k+1][i+j]==1 && s[k]==s[k+1]) dp[i][i+j]=min(dp[i][i+j],dp[i][k]);
                    else if(s[i]==s[i+j] || s[k]==s[k+1] || s[i]==s[k+1] || s[k]==s[i+j]){
                        x=1;
                        dp[i][i+j]=min(dp[i][i+j],dp[i][k]+dp[k+1][i+j]-x);
                    }
                    else dp[i][i+j]=min(dp[i][i+j],dp[i][k]+dp[k+1][i+j]-x);
                }
                cout<<s.substr(i,j+1)<<" : "<<dp[i][i+j]<<endl;
            }
        }
        return dp[0][n-1];
    }
};
23957