What is the time complexity of this code? [closed]?

O(n): outer loop runs logn times, in each iteration: i=1,2,4,8,...n/4 (entrance values) inner loops runs 2*i times (entrance values) so at overall you get: 2+4+8+...+n/2 = n-2 = O(n).

I think this is correct assuming that n is a power of 2 in the first place. If n m^2 then its a little more messy – Jon Egerton Jun 1 at 10:42 3 @Jon: actually there is no mess, in any case, the algorithm will finish iterating when I = n/4 (entrance to last iteration) (and not n/2 as I previously wrote... edited). And is still O(n) – amit Jun 1 at 10:45.

The inner loop executes twice on the first iteration, then four times, then eight times, etc. So you need to figure out where the sum terminates: 2 + 4 + 8 + ... and then work out how to evaluate it (clue: geometric series).

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