After a long time, I returned to TopCoder. I forgot to write algorithm for programming contest such as TopCoder. But previously I realized that it is so important for me to write accurate and fast algorithm within finite time. In order to improve my programming skill again, I returned back to the TopCoder.

SRM is a little hart to me, as first, I tried some practices. Today I solved SRM144 binary code problem.
This problem decode messages recursively. For example, when you get the message `"123210122"`

, this is encode of
`"011100011"`

. Suppose the first message is P, and second is Q. Now below equation is realized.

```
P[i] = Q[i-1] + Q[i] + Q[i+1]
```

With this recusive rule, you have to decode given message. My code is below.

```
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
public class BinaryCode {
public String[] decode(String message) {
// Two answers should be solved
// Each answer is correspond to Q[0] = 0 and Q[0] = 1 case.
String[] ans = new String[2];
Integer start = 0;
Boolean isOut = null;
// Calculate two cases
for (int i = 0; i < 2; i++) {
// In the case of negative value is received, answer should be "NONE"
isOut = false;
start = i;
// For improve speed performance, I use StringBuffer
StringBuffer p = new StringBuffer();
// First and second factor cannot be put on inside loop bacause these are not the sum of three factors
p.append(start.toString());
Integer p1 = Integer.parseInt(message.substring(0, 1)) - Integer.parseInt(p.substring(0, 1));
p.append(p1);
// Decode each digit
for (int j = 1; j < message.length(); j++) {
Integer d = Integer.parseInt(message.substring(j, j+1)) - Integer.parseInt(p.substring(j, j+1)) - Integer.parseInt(p.substring(j-1, j));
if (d < 0) {
ans[i] = "NONE";
isOut = true;
break;
}
p.append(d.toString());
}
// Last digit is not need to retained
if (!isOut) {
ans[i] = p.toString().substring(0, p.length()-1);
}
// This is guard. But I am not satisfied with this line :(
if (message.length() == 1 && (message.charAt(0) == '2' || message.charAt(0) == '3')) {
ans[i] = "NONE";
}
}
return ans;
}
}
```

I wrote this code about 30 minutes. It is not enough to fight on SRM. And in addition to this, I am not satisfied with my algorithm expecially last clause. I donâ€™t want to write exceptional logic as possible. If anyone write code about this problem, please inform me and give me a chance to look into your code.

Thank you.