Algorithms time Complexity

Question

What is the time complexity of this code?

What is the time complexity of the following code?
for (int $\mathrm{i}=\mathbf{1} ; \mathrm{i}<\mathbf{n} ; \mathrm{i}-\mathrm{i} * 3$ ) \{<br /> sum++;
for (int $j=0 ; j sum++;
for (int $k=n-1 ; k>0 ; k--$ )
sum + +;
for (int $\mathrm{h}=\mathrm{n}-1 ; \mathrm{h}>0 ; \mathrm{h}=\mathrm{h} / 2$ )
sum + +;
\} \}
$\mathrm{O}\left(\mathrm{n}^{2}\right)$
$\mathrm{O}\left(\mathrm{n}^{2} \log n\right)$
$\mathrm{O}\left(\mathrm{nlog}^{2} \mathrm{n}\right)$
$\mathrm{O}($ nlogn $)$

Comments

A) ??

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How?

Answer 1

k and h will give n+logn complexity

J , k and h will give n(n+logn) complexity

I , j ,k and h will give

logn base 3(n^2+nlogn) complexity

Finally, n^2 logn base 3 + nlogn lognbase 3

Remove the negligibity and we get o(n^2) as complexity.

Comments

Thanks :)

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I can't properly understand this step

I ,J ,K and H will give $\log_{3}n* (n^2 + n log n)$complexity

and then

I thought it will become  $\log_{3}n* (n^2 )$ + $\log_{3}n* (nlog n )$

which will be O($n^2 log_3 n$) and that equals O($n^2 log n)$.

Please correct me where I'm wrong

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for comparing ,remove the common terms thus logn base3 got cancelled out,remaing is n^2 and nlogn..... most significant term is n^2.