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Zeus2013
Posts: **1**Member

in Algorithms

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!

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!

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## Comments

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

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