On Sun, Aug 7, 2011 at 11:06 PM, Dave <dave_and_da...@juno.com> wrote:
> @Yasir: Sort the numbers. Then
>
> int i = 0, j = 1, m, p = 0;
> while( j < N )
> {
> m = a[j] - a[i];
> if( m = K )
> ++p;
> else if( m < K )
> ++j;
> else
> ++i;
> }
> // answer = p
>
Looks like an infinite loop in there to me...
>
> Complexity = O(n log n).
>
> Dave
>
> On Aug 7, 12:53 pm, Yasir <yasir....@gmail.com> wrote:
> > Guys,
> > Let's try to find out an efficient solution for the following prob:
> >
> > Given N numbers , [N<=10^5] we need to count the total pairs of
> > numbers that have a difference of K. [K>0 and K<1e9]
> >
> > Input Format:
> > 1st line contains N & K.
> > 2nd line contains N numbers of the set. All the N numbers are assured
> > to be distinct.
> >
> > Output Format:
> > One integer saying the no of pairs of numbers that have a diff K.
> >
> > PS: Note that n is very large, so try to come up with an efficient
> > solution. :)
> >
> > Source:http://www.interviewstreet.com/recruit/challenges/dashboard/
>
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--
Shuaib
http://www.bytehood.com
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