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Didnt read the entire lot....but here is what i was thinking..
- gam8it September 12, 2012lets take the first eg: 1220022111001010101020201022210011012
I'd traverse the list...
divide the list into 3 lists...
List 1 : contains all 0
List 2: all 1s
List 3: all 2s
so.... If 1 or 2 is encountered... i'd just remove the node from the original list and add it to the designed list(somthing like a bucket).
for each removal of node from the original List ... i'd keep track of the first and last nodes of the 1's and 2's List.
also do somthing like... node->next = removednode->next
At the end...i'd expect 2 seperate lists of 0's,1's and 2's
I'd just club them together after this..
as I am traversing the original list only once... I guess my complexity should remain O(n).
is the above feasible ?