<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) {
        string sir;
        int dp[105][105];
        for (int i = 0; i < 105; i ++) {
            for (int j = 0; j < 105; j ++) {
                dp[i][j] = -1;
            }
        }
        for (int i = 0; i < s.size(); i ++) {
            if (i == 0 || s[i] != s[i-1]) {
                sir.push_back (s[i]);
            }
        }
        
        return solve (0,sir.size()-1,sir,dp);
    }

    int solve (int a, int b, string sir, int dp[105][105] ) {
        if (a == b) {
            return 1;
        }
        int minim = 1e9;
    
        for (int i = a; i < b; i ++) {
            if (dp[a][i] == -1) {
                dp[a][i] = solve (a,i,sir,dp);
            }
            if (dp[i+1][b] == -1) {
                dp[i+1][b] = solve (i+1,b,sir,dp);
            }
            minim = min (minim, dp[a][i] + dp[i+1][b]);

            if (i!= a && i != b && sir[i] == sir[a] && sir[i] == sir[b]) {
                if (dp[i][b] == -1) {
                    dp[i][b] = solve (i,b,sir,dp);
                }
                minim = min (minim, dp[a][i] + dp[i][b] - 1);
            }
        }

        if (sir[a] == sir[b]) {
            if (dp[a+1][b-1] == -1) {
                dp[a+1][b-1] = solve (a+1,b-1,sir,dp);
            }
            minim = min (minim,dp[a+1][b-1] + 1);
        }

        return minim;
    }
};
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