### big-o

#### Big Oh notation loop manipulation

If I had two nested loops where the outer had a big oh notation of logn and the inner had one of n does that mean the overall notion would be log2n? because since n changes every time the outer loop executes, then the inner loop is technically running logn times but it loops n times. I apologise if this question sounds stupid. This is how the loop looks like: outer loop runs while n>0 inner loops runs n times n=(1/4)n I'm sorry if my formatting is off, I spent a few minutes trying to figure out how to use latex here and couldn't quite crack it

Time complexity is O(n): First time, inner loop iterates n times Second time, inner loop iterates n/4 times Third time, inner loop iterates n/16 times ... K'th time, inner loop iterates n/(4^k) times Summing them up: n + n/4 + n/16 + ... + n/4^k + ... + n/(4^log_4(n)) This is sum of geometric series with: a = n r = 1/4 And it's sums is bounded by: (n/3/4) = 4n/3 accorsing to the formula for r < 1

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