<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:
    string s;
    unordered_map<int, int> memo;
    vector<int> next;
    int dp(int l, int r) {
        if (l >= r) {
            return 0;
        }
        if (memo.find((l << 8) + r) != memo.end()) {
            return memo[(l << 8) + r];
        }
        int k = s[l];
        int ans = 1 + dp(l + 1, r);

        int i = next[l];
        while (i != -1 && i < r) {
            ans = min(ans, dp(l, i) + dp(i + 1, r)); 
            i = next[i];
        }
        memo[(l << 8) + r] = ans;
        return ans;
    }
    int strangePrinter(string s) {
        s.erase(unique(s.begin(), s.end()), s.end());
        this->s = s;
        std::array<int, 26> ch;
        std::fill(ch.begin(), ch.end(), -1);
        vector<int> next(s.size(), -1);
        for (int i = s.size() - 1; i >= 0; --i) {
            if (ch[s[i] - 'a'] != -1) {
                next[i] = ch[s[i] - 'a'];
            } 
            ch[s[i] - 'a'] = i;
        }
        this->next = next;
        return dp(0, s.size());
    }
};
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