T(nlgn) algorithm - Programmers Heaven

#### Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!

#### Categories

Welcome to the new platform of Programmer's Heaven! We apologize for the inconvenience caused, if you visited us from a broken link of the previous version. The main reason to move to a new platform is to provide more effective and collaborative experience to you all. Please feel free to experience the new platform and use its exciting features. Contact us for any issue that you need to get clarified. We are more than happy to help you.

# T(nlgn) algorithm

Posts: 1Member
The question:
Suppose you are given a set P of integers and another integer x. We wish to use a T(nlgn) algorithm to decide whether there are 2 integers in P and the product (multiplication) of these two integers equals to x. Show your algorithm. (You can use pseudo code or by illustration only)

My answer:
Var x;
Function equate(list m)
If(length.m >=1)
Var l.list, r.list;
Var int middle = length.m /2;
For(int i = 0; i < middle; i++)
For(int j = 1;j < middle; j++)
l.list(i)*l.list(j)
if(l.list(i)*l.list(j) =x)
print l.list(i), l.list(j)
For(int i = middle; i < m; i++)
For(int j = middle; j < m; j++)
r.list(i)*r.list(j)
if(r.list(i)*r.list(j) = x)
print r.list(i),r.list(j)

I want to make sure if this is correct! If I have to change something please show me on my code what I have to change, and explain! Thanks!

## Comments

• FrancePosts: 1Member

I don't think your function is correct. Looks like you tried to write a kind of recursive function which does not call itself. And where are you r and l lists filled in ? If it was corrected, I think you would be in n square. Else the solution is pretty simple: sort your list in ascending order (saving the initial indices in each element). Takes nlogn. Then parse the list and if number x(i) divides P, let y = P/x, then seek y in the sorted list, (find j so that x(j)=y), in logn time. So nlogn + n*logn, so all in all it's nlogn

Sign In or Register to comment.