<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 n = s.size();
        vector<vector<int>> memo(n, vector<int>(n, -1));
        return dp(s, 0, n - 1, memo);
    }

private:
    int dp(string& s, int i, int j, vector<vector<int>>& memo) {
        if (i > j) return 0;
        if (memo[i][j] != -1) return memo[i][j];

        int result = dp(s, i + 1, j, memo) + 1;
        for (int k = i + 1; k <= j; ++k) {
            if (s[k] == s[i]) {
                result = min(result, dp(s, i, k - 1, memo) + dp(s, k + 1, j, memo));
            }
        }

        memo[i][j] = result;
        return result;
    }
};
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