<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 dp[1005][1005];
    int strangePrinter(string s) {
        int n = s.size();
        memset(dp,0x7f,sizeof(dp));
        for(int i=0;i<n;i++) dp[i][i] = 1;
        for(int i=1;i<n;i++) dp[i][i-1] = 0;
        for(int i=2;i<=n;i++){
            for(int j=0;j+i-1<n;j++){
                int r = j+i-1;
                dp[j][r] = dp[j][r-1]+1;
                for(int k=j;k<r;k++){
                    if(s[k]==s[r]) dp[j][r] = min(dp[j][r], ((k>0)?dp[j][k-1]:0)+dp[k+1][r]);
                }
            }
        }
        return dp[0][n-1];
    }
};
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