As problem says N is very large, I think sorting is not the right
thing as that would be minimum (n log n) time
how about this
Let's say sum is S
we take an map<int,boo> map and start reading integerts num
if ( num > S ) discard
else {
if ( && map[num] == false){
map[ S - num ] = true ;
} else {
}
On Jun 12, 11:40 am, Chakravarthi Muppalla <[email protected]>
wrote:
> I would use an array.
>
> a[] = 1 3 7 8 9 78 67 23
> a[] = 1 3 7 8 9 23 67 78 //sort the array
> reqSum = 30;
> for i :a.length-1; i>=0; i--
> t = reqSum - a[i];
> if(t is negative) continue;
> find(t);//in the rest of the array;(binary search)
> if(t found in the array) return index of t, current value of i;
> I guess it works.(we may have to use a hash map to speed it up in the long
> run).
>
> On Sat, Jun 12, 2010 at 10:29 AM, Rohit Saraf
> <[email protected]>wrote:
>
>
>
> > I guess it can be done in efficiently using a simple divide and conquer
> > scheme almost imitating mergesort.
> > Can you think of it now? :D
>
> > --------------------------------------------------
> > Rohit Saraf
> > Second Year Undergraduate,
> > Dept. of Computer Science and Engineering
> > IIT Bombay
> >http://www.cse.iitb.ac.in/~rohitfeb14
>
> > On Fri, Jun 11, 2010 at 10:07 PM, sharad kumar
> > <[email protected]>wrote:
>
> >> all possible pairs
>
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> Thanks,
> Chakravarthi.
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