Find Big Oh for nested for loop with floor function

Question

sum $\gets 0$
for $i\gets 1$ to $n$ do
$~~~~$ for $j \gets 1$ to $\left\lfloor\dfrac{n}{i}\right\rfloor$ do
$~~~~~~~~~$ sum $\gets$ sum + ary$[i]$

I know $\sum_{i=1}^{n} = \frac{n(n+1)}{2}$ so Im pretty sure I can just say $O(n^2)$ but Im not sure if this is the answer because perhaps its not precise enough. Im not sure how to incorporate the floor function. I also feel like the sum doesnt go to n it goes to $\left\lfloor\dfrac{n}{i}\right\rfloor$.

score 0 · Answer 1 · edited Apr 13 '17 at 12:19

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Up to the rounding of the upper bound what you are looking for is the sum $n+\frac{n}{2}+\frac{n}{3}+...+\frac{n}{n}$, which is $n$ times a partial sum of the harmonic series. You can find an approximation here, Is there a partial sum formula for the Harmonic Series?, and basically, your complexity will be $O(n\log n)$.

edited Apr 13 '17 at 12:19

Community

1

answered Sep 09 '15 at 13:07

Shahar Even-Dar Mandel

1,158

But would this answer still be consitent with there being a floor function? – yako Sep 09 '15 at 13:24
Yes. The rounding subtracts at most one iteration from each of the inner loops, so the bound is $O(n\log n-n)=O(n\log n)$. – Shahar Even-Dar Mandel Sep 09 '15 at 13:38
The formula Im looking at in my book for harmonic serires says for n>1 then ln n < Hn < ln n + 1. Why are using log instead of ln? – yako Sep 09 '15 at 13:41
For complexity purposes all logarithms are equivalent. – Shahar Even-Dar Mandel Sep 09 '15 at 13:43

Find Big Oh for nested for loop with floor function

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