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
<rohit.kumar.sa...@gmail.com>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
> <sharad20073...@gmail.com>wrote:
>
>> all possible pairs
>>
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Thanks,
Chakravarthi.
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