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I have a simple solution in O(nlogn). First sort the list of numbers. Then iterating from starting (i=1 to n). Do a modified binary search for the remaining n-i+1 elements with l=i ,r=n and find the element of index with floor (k-a[i]). No_of_pairs+= g-i+1.
- kp January 05, 2016Please comment if there's any error with the logic.