<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>> dp(n+1, vector<int>(n+1));
        unordered_map<char, vector<int>> pos;
        for (int i = 0; i < n; i++) {
            pos[s[i]].push_back(i);
            dp[i][i] = 1;
            dp[i+1][i] = 0;
        }
        for (int i = 2; i <= n; i++) {
            for (int j = 0; j <= n - i; j++) {
                dp[j][j + i-1] = 1 + dp[j+1][j+i-1];
                for (int k = 0; k < pos[s[j]].size(); k++) {
                    if (pos[s[j]][k] > j && pos[s[j]][k] <= j + i - 1) {
                        dp[j][j + i-1] = min(dp[j][j+i-1], dp[j+1][pos[s[j]][k]] + dp[pos[s[j]][k]+1][j+i-1]);
                    }
                }
            }
        }
        return dp[0][n-1];
    }
};
13217