<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>

#include <vector>
#include <unordered_map>
#include <algorithm>
#include <string>
#include <utility>  // For std::pair

using namespace std;

class Solution {
public:
    struct hash_pair {
        template <class T1, class T2>
        size_t operator()(const pair<T1, T2>& p) const {
            auto hash1 = hash<T1>{}(p.first);
            auto hash2 = hash<T2>{}(p.second);
            return hash1 ^ hash2;
        }
    };

    unordered_map<pair<int, int>, int, hash_pair> dp;

    int dfs(const string& s, int i, int j) {
        if (i == j) return 1;
        if (i > j) return 0;
        if (i == j - 1) return s[i] == s[j] ? 1 : 2;
        
        if (dp.find({i, j}) != dp.end()) return dp[{i, j}];

        if (s[i] == s[i + 1]) return dp[{i, j}] = dfs(s, i + 1, j);
        if (s[j] == s[j - 1]) return dp[{i, j}] = dfs(s, i, j - 1);

        int minTurns = INT_MAX;
        for (int k = i; k < j; ++k) {
            minTurns = min(minTurns, dfs(s, i, k) + dfs(s, k + 1, j));
        }
        if (s[i] == s[j]) --minTurns;

        return dp[{i, j}] = minTurns;
    }

    int strangePrinter(string s) {
        return dfs(s, 0, s.size() - 1);
    }
};

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