<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<int, unordered_map<int, int>> dp;
public:
    int strangePrinter(string s) {
        return solve(s, 0, s.size() - 1);
    }
    int solve(string& s, int start, int end) {
        if(start == end) {
            return 1;
        } else if(start > end) {
            return 0;
        }
        if(dp[start][end]) {
            return dp[start][end];
        }
        int MIN = INT_MAX;
        for(int i = start; i < end; i++) {
            MIN = min(MIN, solve(s, start, i) + solve(s, i + 1, end));
        }
        return dp[start][end] = MIN - (s[start] == s[end]);
    }
};
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