Basically the index of ( a + b) in the final array will be ceil(( index of a
+ index of b )/2).
Also if there is a clash of indices you just have to compare the values at
those indices and adjust them.
I have run my concept with below two arrays and it works !!!
Arary A: 1 2 3 4 5 6
Array B: 2 3 5 6 8 9
addition of indices 8 4 5 11 6 11
addition of values (2+9) ( 1+5) (4+2) (6+8) ( 3+3) ( 5 + 9)
values: 11 6 6 14 9 14
Added indices: 4 5 6 8 11 11 ( These are not sorted indices, you will
know the indices of values in the final array right away after looking at
the indices of a and b )
indices/2: 2 3 3 4 6 6
corresponding final values 6 6 6 11 14 14
- Kishen Das
On Fri, Apr 30, 2010 at 7:05 AM, divya <sweetdivya....@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.
>
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