<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) {
        int n = s.size();
        function<int(int, int, int)> dfs;
        int f[n][n][27];
        memset(f, -1, sizeof f);
        dfs = [&](int L, int R, int color) {
            if (L > R) {
                return 0;
            }
            if (f[L][R][color] >= 0) {
                return f[L][R][color];
            }
            if (L == R) {
                if (color == s[L] - 'a') {
                    return f[L][R][color] = 0;
                } else {
                    return f[L][R][color] = 1;
                }
            }
            if (s[L] - 'a' == color) {
                return f[L][R][color] = dfs(L+1, R, color);
            }
            if (s[R] - 'a' == color) {
                return f[L][R][color] = dfs(L, R-1, color);
            }
            f[L][R][color] = n;
            for(int i=L;i<=R;i++) if (s[i] == s[L]) {
                f[L][R][color] = min(f[L][R][color], dfs(L+1, i, s[L] - 'a') + dfs(i+1, R, color) + 1);
            }
            // printf("%d %d %d: %d\n", L, R, color, f[L][R][color]);
            return f[L][R][color];
        };
        return dfs(0, n-1, 26);
    }
};
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