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.
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