just address in virtual memory that previously relates to some data

@SilvestrPredko A I said, the address _can_ be reused if the new allocation point "call to new or malloc, ..." points to the same address. If your question is: Is the memory lost until I restart the computer after I call "delete". Then the answer is no, the freed memory returns to the memory allocation pool and has to be requested again.

Exactly. And this doesn't delete any memory at this address. If you call delete and still read at this location you *can* see your own stuff. The memory doesn't change until a new alloc happends and writes new data at this address.

Exactly. And this doesn't delete any memory at this address. If you call delete and still read at this location you *can* see your own stuff. The memory doesn't change until a new alloc happends and writes new data at this address.

oh it just set this address like as freed, same behaviour as file are deleted from hard disk. Just remove record about it

yes

Exactly

thanks guys)

u r welcome :)

Anybody expert in Algorithms?

What will be the time complexity of T(n)=T(n-1)+T(n-2)+n

What will be the time complexity of T(n)=T(n-1)+T(n-2)+n

O(!n) I think

O(!n) I think

How???

assuming the naive implementation

Anyone to send me a good c compiler

assuming the naive implementation

Ohkkk

Anyone to send me a good c compiler

ultimate kek

Anyone to send me a good c compiler

G - Good C - C C - Compiler

What will be the time complexity of T(n)=T(n-1)+T(n-2)+n

A smart implementation would be O(n) tho

ultimate kek

Can I get link to download please 🤲

A smart implementation would be O(n) tho

Ohkk

I think linear -> O(n) @ritu_raj_s

Can I get link to download please 🤲

Ever heard of google?

It's a website, many people use it

Ever heard of google?

Issokay 👊

I think linear -> O(n) @ritu_raj_s

O(an)

I think linear -> O(n) @ritu_raj_s

I don't know why I am getting O(n^2)

Can you provide me soln???

@linuxer4fun : what is "a" in O(an) ? I'm not an expert in determining time complexity

its a coefficient

If you expand it, T(n) = n +sigma i = 1 to i = n i * (n - i) = n + O(n ^ 2) - O(n ^ 3) so should be O(n^3) I think

I don't know why I am getting O(n^2)

For the naive one: You call for each step n-1 times the thingy + n-2 times it recursively, so you make a factorial complexity. For the linear one, you just have a loop and a store and go increasingly for higher values so it will always be linear to n

@akki: Hmm... perhaps because it is a recursive function which is results as the same or similar to a nested loop. two nested loops have the complexity of O(n²)

@linuxer4fun : If a is a coefficient, then should be the complexity then still be linear O(n)? Just wondering

@akki: Hmm... perhaps because it is a recursive function which is results as the same or similar to a nested loop. two nested loops have the complexity of O(n²)

a self recursive function would have a complecity of (!n) like for example the naive fibonacci

its a coefficient

coefficients get ignored

@linuxer4fun : If a is a coefficient, then should be the complexity then still be linear O(n)? Just wondering

linear is always a coefficient

coefficients get ignored

Oh

It's common ^^

linear is always a coefficient

you don't write the coeff normally

For the naive one: You call for each step n-1 times the thingy + n-2 times it recursively, so you make a factorial complexity. For the linear one, you just have a loop and a store and go increasingly for higher values so it will always be linear to n

Ohkk......

@akki: Hmm... perhaps because it is a recursive function which is results as the same or similar to a nested loop. two nested loops have the complexity of O(n²)

👍👍👍👍👍

you don't write the coeff normally

:(

linear is always a coefficient

unless you need to go through the steps twice or thrice explicitly and want to compare on that basis

If you expand it, T(n) = n +sigma i = 1 to i = n i * (n - i) = n + O(n ^ 2) - O(n ^ 3) so should be O(n^3) I think

Different one

What will be the time complexity of T(n)=T(n-1)+T(n-2)+n

Depends on the end condition too what is T(0)

@linuxer4fun : Hmm... you just gave me some homework to do :)

@linuxer4fun : Hmm... you just gave me some homework to do :)

Sure ;)

unless you need to go through the steps twice or thrice explicitly and want to compare on that basis

I see

But for T(n-1)+n

I am getting O (n^2)

Is it correct or incorrect?

This should be correct

The one you mentioned earlier is different

The one you mentioned earlier is different

For the previous one I also think answer is O(n^3)

What the wrong with this ?

everything.

Why ?

everything.

😂😂😂

btw: who is blowing your friend who is in the background?

Plz help not comment

Plz help not comment

i'm not even going to try helping you

i'm not even going to try helping you

Can't be bothered to even try reading that tbh

Would be a waste of time

Can't be bothered to even try reading that tbh

you should, its the perfect "try not to cringe" compilation

you should, its the perfect "try not to cringe" compilation

Lmao yes

yea

Hello every one

you mean the slave in the background?

you mean the slave in the background?

Me as I read that

xDD

Lel, a few seconds ago our trainee asked me if I could explain him some code in one of our libraries. I came to his desk, he showed me two files full of TMP. Some of them written by me 😆I just said "Uhm.. This is a special topic of C++, you don't need to work on that code, fix some other bugs." - explaining a beginner how TMP code works is like black magic for him

TMP?

Template meta programming

my favourite topic

templates allow for a lot of black magic

template magic :D

Yee, pretty usefull for some stuff and libs, but absolute overkill for beginners

xDDD