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I've been trying this problem but have run into something that confuses me. The examples provided:
input: 3 4 S3 D2 D3 S2 4 S3 D2 S2 D3 6 S3 D2 D3 S2 S4 D4
answer: 3j 1 1 0 5j 4 3j 2 1
According to the text the second problem=> 4 S3 D2 S2 D3 is not solvable but it seems to me that it is using the same 4 moves as the first problem => 4 S3 D2 D3 S2
starting with S3 D2 S2 D3 => do 3j yields => D3 D2 S2 -. do 1 yields -> D2 S2 do 1 yields -> S2 => a correct solution ???
Am I missing something or is it just a type?
Thanks
Hi,
0 result means only skipping the case (as you have to solve at least 5 cases, not all of them, you can skip any cases you want) 0 does not mean that the case is unsolvable. There are many solutions can exist to a case. Your example above is right, the second case can be solved by what you wrote.
Thanks I missed that.