<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 strangePrinter(string s) {
        regex re("(.)\\1+");
        s = regex_replace(s, re, "$1");

        int n = s.size();
        vector dp(n, vector<int>(n, n));

        for (int length = 1; length <= n; length++) {
            for (int l = 0; l <= n - length; l++) {
                int r = l + length - 1;
                int j = -1;
                for (int i = l; i < r; i++) {
                    if (s[i] != s[r] && j == -1) {
                        j = i;
                    }
                    if (j != -1) {
                        dp[l][r] = min(dp[l][r], 1 + dp[j][i] + dp[i + 1][r]);
                    }
                }

                if (j == -1) {
                    dp[l][r] = 0;
                }
            }
        }

        return dp[0][n - 1] + 1;
    }
};
16797