620 - Cellular Structure

[mamun]

Thanks. I understand it now. I'll give it a try.

[Psyco]

I think this problem's tip is..

len=string length

Mutant : len is 1, and str[0]='B'

Simple : if len is 1, and str[0]='A'

Fully-Grown : str[len-1]='B', and str[len-2]='A'

Mutagenic : str[0]='B', and str[len-1]='A'

Else : Mutant



In case of Mutant, it has 2 cases to print.

I got AC thie problem in C, line 18.

[jan_holmes]

How could "BAAB" become "MUTANT" ??? Shouldn't it be "FULLY-GROWN" ??? because in the problem statement :

fully-grown stage O = OAB


and in this case, "BAAB", BA = O, and AB,so BAAB should be "FULLY-GROWN"

Thx...

[Raiyan Kamal]

"BAAB", BA = O, and AB,so BAAB should be "FULLY-GROWN"

How do you know BA = o ?

BA does not match with any of the given patterns.

Code: Select all

A <- simple
OAB <- fully-grown
BOA <- mutagenic

So it must be treated as 'mutant'. And since the input string has a substring who is mutant, the entire string is actually mutant.