<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:
    int memo[105][105];

    int foo (int start, int end, string s) {
         if (start > end) return 0;
        if(memo[start][end] != -1) return memo[start][end];
        int ans = 1 + foo(start+1, end, s);
        for(int k=start+1; k<=end; k++){
            if(s[start] == s[k])
                ans = min(ans, foo(start+1, k-1, s) + foo(k, end, s));
        }
        return memo[start][end] = ans; 
    }

    int strangePrinter(string s) {
        string temp = "";
        for(int i=0; i<s.size(); i++) {
            temp += s[i];
            while(i < s.size()-1 && s[i] == s[i+1]) i++;
        }
        s = temp;
        memset(memo, -1, sizeof(memo));
        return foo(0, s.size()-1, s);
    }
};
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