You can use policy based data structure ( Ordered set ) . Then you can traverse the array from the right and keep checking no. of elements using order_of_key(a[i]) . o(nlogn) tc

Also ig it can be done using Divide and conquer ( Merge sort )

Here similar problem on leetcode :

https://leetcode.com/problems/count-of-smaller-numbers-after-self/description/

You can also use Segment Tree, if array value is large in that case you have to normalize it down to <=n range.

you can sort the array in decreasing order keeping their index then you build segment / fenwick tree ans for index i is i — sum(0..j) j denotes index of element in original array then update the value of segment / fenwick tree. sum(0..i) denotes cnt of elements having value less <= element at index i and index < j

Caculate i < j and a_i < a_j. Enumerate from n to 1. For position i, you only need to caculate how many greater than a_i. Use BIT, query(n) — query(a[i]) is answer for position i.

i guess find the next greater element for every element to its right and add 1 to the answer* of that element to get the answer for this element.

• answer means-- number of elements to its right such that they are greater than this element

I guess you are trying to find count of inversions in array which is

if i < j and a[i] > a[j]

This can be done using merge sort along with a simple modification.

Link to understand sol

Statement's really unclear, can you clarify what the problem is?