<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:
    unordered_map<string, int> memo;

    int helper(string& s, int left, int right) {
        if (right - left <= 1) return right - left;

        string key = to_string(left) + "," + to_string(right);
        if (memo.find(key) != memo.end()) return memo[key];

        if (s[left] == s[left+1] || s[left] == s[right-1]) return memo[key] = helper(s, left+1, right);
        if (s[right-2] == s[right-1]) return memo[key] = helper(s, left, right-1); 

        int result = INT_MAX;
        for (int i = left+1 ; i < right ; i++) {
            result = min(result, helper(s, left, i) + helper(s, i, right));
        }

        return memo[key] = result;
    }

    int strangePrinter(string s) {
        return helper(s, 0, s.length());
    }
};

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