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On step 4 you should not increment k, but increase it by some factor (for example 2). Then performance will be linear in N - the length of the train. For scale factor 2 worst case performance is when N = 2^M+1, you would travel back and force total distance of sum(2*k)=sum(2*2^m), for m=1,(M+1), which is approximately 2*2^(M+2) = 8*2^M ~ 8*N.
- anonymous November 06, 2015