Not a triangle

practice

#1

I am getting WA for the question:http://www.codechef.com/problems/NOTATRI/
Anyone plz. explain where i am wrong.Thnx.
My code:
http://www.codechef.com/viewsolution/2213497


#2

sorting will help in terms of time. after sorting, start from the greatest value of length, if sum of smallest value and value lesser than(l[k-1]) taken value(l[k]), then add difference to count(cnt=cnt+l[k-1]-l*), because sum of all indices between those numbers, if considered particularly, will be lesser than taken value(l[k]). if sum of those two particular indices is greater than or equal to taken value(l*+l[j]>=l[k]), then increment the smallest index(i++). loop the aforementioned case for all index. voila.


#3

1.What we will be doing is , we will take a set of 3 lengths say {first,second,third} such that first<=second<=third. This way we will count only unique sets.
2.After this we will fix first=0 and third = n-1 , i.e first and last elements of sorted array.Yes , array has to be sorted . Quick sort or Merge sort can be used.
3.Now loop is on third = n-1 to 0. for each chosen third , i have inner loop on first.
4.Now second begins from third-1 and move towards first. Now if we find some second such that first + second < third then our condition is satisfied and all such numbers between first and second are suitable candidates for “second”. hence we count their number and add them.

For a better understanding of solution look here