<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 f(string s, int i, int j, vector<vector<int>>&dp)
    {
        if (i==j)
        {
            return 1;
        }

        if (dp[i][j] != -1)
        {
            return dp[i][j];
        }

        int mi = 10000;
        for (int z = i;z<j;z++)
        {
            int a = 0;
            int b = 0;
            if (dp[i][z] != -1)
            {
                a = dp[i][z];
            }
            else
            {
                a = f(s, i, z, dp);
                dp[i][z] = a;
            }
            if (dp[z+1][j] != -1)
            {
                b = dp[z+1][j];
            }
            else
            {
                b = f(s, z+1, j, dp);
                dp[z+1][j] = b;
            }
            mi = min(mi, a + b);
        }
        if (s[i] == s[j])
        {
            mi--;
        }
        return dp[i][j] = mi;
    }

    int strangePrinter(string s) 
    {
        int n = s.size();
        vector<vector<int>> dp(n+1, vector<int>(n+1, -1));
        int ans = f(s, 0, n-1, dp);
        return ans;
    }
};
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