<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 {
    unordered_map<long long, int> cache;
public:
    int strangePrinter(const string &s) {
        return print(s, 0, static_cast<int>(s.length()) - 1);
    }

    int print(const string &s, const int l, const int r) {
        if (l > r) {
            return 0;
        }
        if (l == r) {
            return 1;
        }
        long long key = (static_cast<long long>(l) << 31) + r;
        auto resultPos = cache.find(key);
        if (resultPos != cache.end()) {
            return resultPos->second;
        }
        int result = s[l] == s[r] ? print(s, l + 1, r) : r - l + 1;
        for (int i = l; i < r; ++i) {
            result = min(result, print(s, l, i) + print(s, i + 1, r));
        }
        cache[key] = result;
        return result;
    }
};
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