<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 n=0;
    int strangePrinter(string s){
        n=s.size();
        unordered_map<int,unordered_map<int,int>> dp;
        if(n==1){
            return 1;
        }
        for(int i=0;i<n;i++){
            dp[i][i]=1;
        }
        int j=0;
        for(int l=2;l<n+1;l++){
            for(int i=0;i<n-l+1;i++){
                j=i+l-1;

                dp[i][j]=dp[i+1][j]+1;
                for(int k=i+1;k<j+1;k++){
                    if(s[i]==s[k]){
                        if(j>k){
                            dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]);
                        }
                        else{
                            dp[i][j]=min(dp[i][j],dp[i][k-1]);
                        }
                    }
                }
            }
        }
        return dp[0][n-1];
    }
};
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