<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 strangePrinter(string s) {
        int dp[s.size() + 1][s.size() + 1];
        for (int i = 0; i <= s.size(); i++){
            for (int j = 0; j <= s.size(); j++){
                dp[i][j] = INT_MAX - 10;
                if (j < i){dp[i][j] = 0;}
            }
        }
        for (int i = 0; i <= s.size(); i++){
            dp[i][i] = 1;
        }
        for (int j = 0; j < s.size(); j++){
            for (int i = j - 1; i >= 0; i--){
                dp[i][j] = dp[i + 1][j] + 1;
                for (int l = i + 1; l <= j; l++){
                    if (s[i] == s[l]){
                        dp[i][j] = min(dp[i][j], dp[i][l - 1] + dp[l + 1][j]);
                    }
                }
            }
        }
        return dp[0][s.size() - 1];
    }
};
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