<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 solve(string s , int i , int j , vector<vector<int>>&dp)
    {
        if(s.size() == 0 || i>j)
        {
            return 0 ;
        }
        if(i==j)
        {
            return 1 ;
        }
        if(dp[i][j] != -1)
        {
            return dp[i][j] ; 
        }
        int ans = INT_MAX;
        if(s[i]==s[j])
        {
            ans = solve(s , i , j-1 , dp);
        }
        for(int k=i ; k<j ; k++)
        {
            int l , r;
            if(dp[i][k] != -1)
            {
                l = dp[i][k]; 
            }
            else 
            {
                l = solve(s , i , k , dp);
                dp[i][k] = l ;
            }
            if(dp[k+1][j] !=-1 )
            {
                r = dp[k+1][j] ; 
            }
            else 
            {
                r = solve(s , k+1 , j , dp) ; 
                dp[k+1][j] = r ;
            }
            ans = min(ans , l+ r);
        }
        return dp[i][j] = ans ;
    }
    int strangePrinter(string s) 
    {
        int n = s.size();
        vector<vector<int>>dp(n+1 , vector<int>(n+1 , -1));
        return solve(s , 0 , n-1 , dp);  
    }
};
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