<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:
    vector<vector<int>>dp=vector<vector<int>>(101,vector<int>(101,-1));     

    int solve(int i,int j,string &s){
        if(i==j)return 1;
        if(dp[i][j]!=-1)return dp[i][j];

        int turns=1e9;
        for(int k=i;k<j;k++){
            turns=min(turns,solve(i,k,s)+solve(k+1,j,s));
        }
        if(s[i]==s[j])turns--;
        return dp[i][j]=turns;

    }
    int strangePrinter(string s) {
        return solve(0,s.size()-1,s);
    }
};

// Time Complexity : O((101^2)*101) => O(100^3) => O(N^3); 
// Space Complexity: O(101*101) + O(100) => 100^2 => O(N^2);
// Max piles size is 100 (max N);
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