@Coolfrog$ and Gene: I understand now, and agree that my algorithm is
incorrect. Just because I've used a flight in one trip doesn't mean
that it can't also be used in another more expensive trip. Sorry.
Dave
On Oct 17, 10:24 pm, "coolfrog$" <dixit.coolfrog.div...@gmail.com>
wrote:
> @Dave
> yes im also trying to say the same point as Gene
> these problem is similar to
> "*Least fare for a return trip Algorithm*" where u have suggest
> the exactly same algorithm..
> @Gene do see these similar problem
> "[algogeeks] Least fare for a return trip Algorithm"
> here we have to find minimum n pairs sum instead of maximum.
>
>
>
>
>
> On Mon, Oct 18, 2010 at 6:01 AM, Gene <gene.ress...@gmail.com> wrote:
> > Hmm...
>
> > First, drawing the table is O(n^2). So we have to assume you never
> > really draw the table.
>
> > Second, your description is too imprecise to say if it will work in
> > general. Give a pseudocode, please. I don't believe that any
> > algorithm based on local inspection of this table will work. In
> > general you'll be hopping around the table in a wild order.
>
> > Third, I'm not sure why you'd pick an example where a and b are the
> > same array. All the interesting cases are when they're different.
>
> > On Oct 16, 6:35 pm, ligerdave <david.c...@gmail.com> wrote:
> > > like i said before, draw a table w/ x=a and y=b
>
> > > come out w/ the matrix and you would see a patten
>
> > > 30 29 4 3 2 1
>
> > > 30 60 59 34 33 32 31
>
> > > 29 59 58 33 32 31 30
>
> > > 4 34 33 8 7 6 5
>
> > > 3 33 32 7 6 5 4
>
> > > 2 32 31 6 5 4 3
>
> > > 1 31 30 5 4 3 2
>
> > > you start from a[0] and b[0], compare them, take the bigger one, set
> > > the smaller one(for next iteration) and set a limit.
>
> > > for instance, in this case, either one works. let's say a[0] is
> > > bigger, the limit will be a[1]+b[0]. limit is always the element next
> > > to the bigger one in array, plus the biggest in another array.
>
> > > you loop through a[0]+b[i] where i=0 to length of b. stop when the
> > > outcome is less than limit.
>
> > > now you take what is stored(the smaller one) to start the next
> > > iteration(steps above)
>
> > > On Oct 15, 7:56 pm, Gene <gene.ress...@gmail.com> wrote:
>
> > > > Hi Arun. Last time we discussed this problem I came up with the same
> > > > idea. Unfortunately it doesn't work. The problem is that in order
> > > > for the "merge" to be correct, each pair of pointers must be
> > > > guarenteed to produce sums that are in non-increasing order. They
> > > > don't. For example, if you run your program with
>
> > > > int a[N] = { 1, 2, 3, 4,29,30};
> > > > int b[N] = { 1, 2, 3, 4,29,30};
>
> > > > It will produce
>
> > > > (30,30)=> 60 (30,29)=> 59 (29,30)=> 59 (30,4)=> 34 (4,30)=> 34
> > > > (30,3)=> 33
>
> > > > This is wrong because the 4th pair should be (29,29) => 58.
>
> > > > In fact, niether here nor in the previous discussion did we ever get a
> > > > correct solution.
>
> > > > If you figure it out, please post.
>
> > > > On Oct 14, 5:54 am, arun raghavendar <arun.s...@gmail.com> wrote:
>
> > > > > Start merging A and B from the tail end for N elements, just like the
> > way
> > > > > you would do it for a merge sort but with a different constraint
> > based on
> > > > > the sum a[i] and b[i]
>
> > > > > This should work for any value of N, I just hard-coded it for
> > simplicity.
>
> > > > > #include<stdio.h>
> > > > > #define N 6
> > > > > struct OutputType {
> > > > > int a;
> > > > > int b;
> > > > > int value;
>
> > > > > };
>
> > > > > int main() {
> > > > > int a[N] = {1,8,13,24,25,30};
> > > > > int b[N] = {5,6,17,28,29,29};
> > > > > struct OutputType s[N];
> > > > > int i, a_candidate_1 = N-1, a_candidate_2=N-2,
> > b_candidate_1=N-1,
> > > > > b_candidate_2=N-2;
> > > > > s[0].a = a[N-1];
> > > > > s[0].b = b[N-1];
> > > > > s[0].value = a[N-1] + b[N-1];
> > > > > for (i=1;i<N;i++) {
> > > > > if ((a[a_candidate_1]+b[b_candidate_2]) >=
> > > > > (a[a_candidate_2]+b[b_candidate_1])) {
> > > > > s[i].a = a[a_candidate_1];
> > > > > s[i].b = b[b_candidate_2];
> > > > > s[i].value =
> > a[a_candidate_1]+b[b_candidate_2];
> > > > > b_candidate_2--;
> > > > > if (b_candidate_2 < 0) { a_candidate_1--; }
> > > > > } else {
> > > > > s[i].a = a[a_candidate_2];
> > > > > s[i].b = b[b_candidate_1];
> > > > > s[i].value =
> > a[a_candidate_2]+b[b_candidate_1];
> > > > > a_candidate_2--;
> > > > > if (a_candidate_2 < 0) { b_candidate_1--; }
> > > > > }
> > > > > }
>
> > > > > for (i=0;i<N;i++) printf("(%d,%d)=>%3d ", s[i].a, s[i].b,
> > > > > s[i].value);
>
> > > > > return 0;
>
> > > > > }
>
> > > > > -Arun
>
> > > > > On Thu, Oct 14, 2010 at 1:25 PM, Harshal <hc4...@gmail.com> wrote:
>
> > > > > > Given two sorted postive integer arrays A[n] and B[n] (W.L.O.G,
> > let's
>
> > > > > > say they are decreasingly sorted), we define a set S = {(a,b) | a
> > \in A
> > > > > > and b \in B}. Obviously there are n^2 elements in S. The value of
> > such
> > > > > > a pair is defined as Val(a,b) = a + b. Now we want to get the n
> > pairs
>
> > > > > > from S with largest values. The tricky part is that we need an O(n)
> > > > > > algorithm.
>
> > > > > > --
> > > > > > Harshal Choudhary,
> > > > > > III Year B.Tech Undergraduate,
> > > > > > Computer Engineering Department,
> > > > > > National Institute of Technology Surathkal, Karnataka
> > > > > > India.
>
> > > > > > --
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