Your comment brings to mind that we could do an O(n) sort such as a
Radix Sort. Any ordinary implementation will be O(n) as a fixed number
of passes will be required based on the word size.
Dave
On Sep 23, 11:21 am, nishaanth <nishaant...@gmail.com> wrote:
> i dont think there exists a o(n) solution for this
> the best we can do is sort in o(nlogn) and then do a o(n) traversal
>
>
>
>
>
> On Thu, Sep 23, 2010 at 8:48 PM, Dave <dave_and_da...@juno.com> wrote:
> > @Yellow: Change the 12 in a[1] in your test case to 35 and see what
> > you get.
>
> > Dave
>
> > On Sep 23, 10:09 am, Yellow Sapphire <pukhraj7...@gmail.com> wrote:
> > > I hope this will work. It finds the two minimum numbers and then prints
> > the
> > > difference.
>
> > > #include <stdio.h>
> > > #include <stdlib.h>
> > > void find_two_mins(int array[], int size)
> > > {
> > > int i,t;
> > > int min1=array[0], min2=array[1];
> > > for (i=2; i<size; i++){
> > > t=array[i];
> > > if(t<=min1) {
> > > min2=min1;
> > > min1=t;
> > > continue;
> > > }
> > > }
> > > printf("\nMin elements are 1: %d 2: %d", min1,min2);
> > > printf("\n Absolute difference between two minimum elements are %d",
> > > abs(min1-min2));
>
> > > }
>
> > > int main()
> > > {
> > > //int array[10]={ 9,2,3,4,1,7,5,6,8,0};
> > > //
> > > int array[10]={99,12,45,33,88,9098,112,33455,678,3};
> > > find_two_mins(array,10);
> > > return 0;
>
> > > }- Hide quoted text -
>
> > > - Show quoted text -
>
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> --
> S.Nishaanth,
> Computer Science and engineering,
> IIT Madras.- Hide quoted text -
>
> - Show quoted text -
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