@Harshal: your terminating condition would be -
lets say we have set the pointers to index 0 of each to get the min
distance.
for index 0 set the min_dist overall to the max distance among the 3
pairs. Now increase the pointer with the minimum value and check the
max distance between pairs. If distance now is more than min_dist we
terminate and we have the answer
else continue incrementing the pointer with the minimum value until we
get a distance value greater than the previously computed min_dist
value.
correct me if i have misinterpreted ur algo.

On Jun 17, 10:08 pm, Harshal <[email protected]> wrote:
> I think this will work,
> have 3 pointers p,q,r pointing last elements of the 3 lists.
> compute the difference between each pair.
> decrement the index of the list having the min element.
> (at each stage, save the current indices and current max distance).
>
> Same logic for the min distance part, just start from index 0.
>
>
>
>
>
> On Fri, Jun 17, 2011 at 9:43 PM, Ashish Goel <[email protected]> wrote:
> > merge two and if required third  array keeping array tag with the elements
> > walk over the merged list and see adjacent distance which is minimum with
> > the condition that the tage of the adjacent elements are different
>
> > Best Regards
> > Ashish Goel
> > "Think positive and find fuel in failure"
> > +919985813081
> > +919966006652
>
> > On Fri, Jun 17, 2011 at 9:36 PM, Dumanshu <[email protected]> wrote:
>
> >> U have got 3 sorted arrays A1 A2 and A3 having m n and p elements
> >> respectively. A gap of 3 arrays is defined to be max distance between
> >> 3 nos if they are put on a no line say u pick three 2 12 and 7 then
> >> the gap is 10. Now u have to find an efficient way of chosing 3 nos
> >> from these 3 seperate arrays (A1, A2, A3) such that their gap is
> >> minimum. Of course if a num say 2 occurs in all 3 then gap is 0!!!
>
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> Harshal Choudhary,
> III Year B.Tech CSE,
> NIT Surathkal, Karnataka, India.- Hide quoted text -
>
> - Show quoted text -

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