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We can start inquiring the the numbers at indices at 2, 4, 8, ... till we find a 1 at say 2^(n-1) and a 0 at 2^n. Now we can do binary search between these two indices for 0.

We can't apply binary search as we don't know the size of the array and I guess there is no better solution that what's posted by Anonymous.

The bit operators are faster.
So just keep XOR-ing with 1 all the indices starting from the first index.. The index where this operation gives a result 1 is the required index

WHy not use Xors with exponentiating indices as suggested earlier?

'\u2202'

I think the above suggested solution would take no less average time then applying the binary search algorithm . So we have a easy solution for this doing this in 'log n' time . Do u need it further faster than this, even I would like to know if there is any way to improve time complexity .