Compare each array $i$ member to all array $j$ member where $(i > j)$ , in time less than $O(n^2)$
Create a binary tree containing the points, either each node also containing the number of elements in each sub tree . For each candidate, determine the number of candidates arriving earlier, and having same or higher score using the tree, then add the candidates score to the tree. 

Should work in O(n log n). I doubt O(n) is possible. 

PS chi’s answer is a lot easier, and you probably have a library for the sorting, which is the hard part.