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1 little optimization is possible
- prasanna.raj August 21, 2008We need 2 numbers 1 <= 20/2 and 1 >= 20/2.
So we take the list and sort it.
Do a binary search on 20/2 = 10 (or more generically lower bound n/2)
and lets say you have ended up in location k (if k =0 or n-1) u can stright
away say you dont have any elements that sum up to 20
and have 2 pointers 1 running from 0->k and from k->n-1. summing up and checking
if equals 20