Time complexity of a function with while loop

Question

What is the time complexity of the following procedure?

for $x \in \{1,\ldots,n\}$:
$\quad$ $i \gets \lfloor n/2 \rfloor$
$\quad$ while $i \neq x$:
$\quad\quad$ if $i > x$ then $i \gets i -1$, otherwise $i \gets i + 1$.

According to me, the time complexity is $O(n^2)$. The inner while loop starts at $n/2$ and moves towards the value of $x$, worst case it runs $n/2$ times, and best case it runs $0$ times.

What is the exact time complexity of this procedure?

I suggest following the systematic methodology there, then editing your question to show your work. That's how you can verify your intuition. — D.W., Apr 10 '21 at 06:30
Please, post text as text, not as photographs of text. This is not a website for photographers who want to critique your use of color and perspective. It should be possible to copy&paste your text. It also makes your question unreadable to the blind and visually impaired. It also makes it impossible to index both by Stack Exchange's own search engine as well as Google, Bing, and co. — Jörg W Mittag, Apr 10 '21 at 11:44

grus · Answer 1 · 2021-04-10T16:20:35.360

1

Assume $\frac{N}{2}\in\mathbb{N}$, $\frac{N}{2}+(\frac{N}{2}-1)+\cdots+1+0+1+\cdots +(\frac{N}{2}-1)=\frac{N}{2}+2\times\sum_{k=1}^{N/2-1}k=\frac{N}{2}+\frac{N^2-2N}{4}=\frac{N^2}{4}\in O(N^2)$

edited Apr 10 '21 at 16:20

answered Apr 10 '21 at 15:11

grus

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That equation cannot be true (although the sums is indeed in $O(N^2$). – Steven Apr 10 '21 at 15:58
@Steven I used - instead of + by mistake, I edited it and wrote it more precisely. Are there any other error here? – grus Apr 10 '21 at 16:21

score 0 · Answer 2 · answered Apr 13 '21 at 08:24

The while loop runs for $|x - \lfloor n/2 \rfloor|$ many iterations. Therefore the running time is proportional to $$ \sum_{x=1}^n |x - \lfloor n/2 \rfloor|. $$ To ease the calculation a bit, let us assume that $n$ is even. We break the sum into two parts: $$ \sum_{x=1}^{n/2} (n/2 - x) + \sum_{x=n/2+1}^n (x - n/2) = \sum_{x=0}^{n/2-1} x + \sum_{x=1}^{n/2} x = \frac{(n/2-1)(n/2)}{2} + \frac{(n/2)(n/2+1)}{2} = \Theta(n^2). $$

Time complexity of a function with while loop

2 Answers2