Hi,

ah. Ok. I understand now your particular issue. Yes. Your approach is 
perfectly feasible. By paying the price of calling f multiple times for an 
element and realising each group twice you can make the two sequences 
(take-while and drop-while) independent and hence don't retain the head of 
the first group when realising the second group. With the count solution 
the first group would have been also realised and kept in memory. However 
since the take and drop are independent the first group is realised, but 
thrown away immediatelly. 

Your posted solution should work fine.

Meikel

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