O 500-999: Fill & Download for Free

It’s not possible. Accessing an arbitrary element in a linked list is [math]O(n)[/math]. You blow your big-Oh budget on the first step of the binary search.Binary search is considered [math]O(\lg n)[/math] if the the container is sorted, and you can access arbitrary elements in the container in constant time. Linked lists fail the second criterion.Consider a linked list with 1000 elements, numbers 1 to 1000:[math]1 \leftrightarrow 2 \leftrightarrow 3 \leftrightarrow \cdots \leftrightarrow 999 \leftrightarrow 1000 \leftrightarrow \square[/math]The first step of the binary search requires you to get to the middle element. This requires 500 steps through the list. (Fun fact: You don’t need to know the list’s length a priori to find the middle element. Exercise for the reader: Why?)Let’s say we’re searching for 1000, and we stick to the binary search rule of only examining midpoints of the segments that remain after each step. We’ll take500 steps to get to 500250 steps to get to 750125 steps to get to 87562 steps to get to 937…In the end, we’ll take 1000 steps to get to 1000. There’s not much way around it. In the end, it’ll be [math]O(n)[/math] steps.The only thing you save over linear search is comparisons: You’ll only need [math]O(\lg n)[/math] comparisons. [EDIT: As Jesse Pollard points out in the comments, this statement only applies to key comparisons. Each step through a linked list still involves a pointer comparison, and that remains [math]O(n)[/math].]

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correction I believed o Owen’s answer too quicklythere was a mistake in Owens answer the last 9 term are indeed each = to 1 but the first 9 are i, -1, -i,1 ,i, -1, -i,1, i which sum to -i explained belowi^1= ii^2 = -1i^3=-1ii^4= 1then it repeats so if you dive the exponent by 4 and there is no remainder the value is 11. If remainder is 1 the value is i is remainder -3 the value is -i and if the remainder is 0 the value is 1 hence the answer is 9 -i