<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>

int dp[101][101];
int solve(int i,int j,string s){
    if(i>j)return 0;
    if(i==j)return 1;
    if(dp[i][j]!=-1)return dp[i][j];
    int ans=INT_MAX;
    ans=min(ans,1+solve(i+1,j,s));
    for(int k=i+1;k<=j;k++ ){
        if(s[k]!=s[i])continue;
        ans=min(ans, solve(i+1,k-1,s)+solve(k,j,s));
    }
    return dp[i][j]= ans;
}

class Solution {
public:
    int strangePrinter(string s) {
        memset(dp,-1,sizeof dp);
        string temp="";
        char c='0';
        for(auto e:s){
            if(e==c)continue;
            temp.push_back(e);
            c=e;
        }
        // temp=s;
        // cout<<temp<<endl;
        return solve(0,temp.length()-1,temp);

    }
};
69602