Count smaller primes on the right of each array element

Given an array A[] of size N, the task for each array element is to count the array elements on their right that are smaller than it and are prime.

Examples:

Input: N = 10, A[] = {5, 5, 17, 9, 12, 15, 11, 7, 39, 3} 
Output: 2 1 3 2 3 3 2 1 1 0 
Explanation: 
For i = 1, elements at j = [2, 10] are valid answer. 
For i = 2, elements at j = [10] are valid answer. 
For i = 3, elements at j = [7, 8, 10] are valid answer. 
For i = 4, elements at j = [8, 10] are valid answer. 
For i = 5, elements at j = [7, 8, 10] are valid answer. 
For i = 6, elements at j = [7, 8, 10] are valid answer. 
For i = 7, elements at j = [8, 10] are valid answer. 
For i = 8, elements at j = [10] are valid answer. 
For i = 9, elements at j = [10] are valid answer. 
For i = 5, no elements are on its right. 

Input: N = 6, A[] = {43, 3, 5, 7, 2, 41} 
Output: 5 1 1 1 0 0
Explanation: 
For i = 1, elements at j = [2, 3, 4, 5, 6] are valid answer. 
For i = 2, elements at j = [5] are valid answer. 
For i = 3, elements at j = [5] are valid answer. 
For i = 4, elements at j = [5] are valid answer. 
For i = 5, no valid answer. 
For i = 6, no valid answer.

Naive Approach: The simplest approach to solve this problem is to traverse the array and for each element, A[i], iterate over all the elements on its right and count the number of elements smaller than A[i] and are prime.

Below is the implementation of the above approach:

5 1 1 1 0 0 

Time Complexity: O(N²)
Auxiliary Space: O(N)

Efficient Approach: The idea is to observe that the above approach can be optimized by iterating over the array from right to left and calculate the required count of primes for each array element using Fenwick Tree. Follow the steps below to solve the problem:

• Traverse the array from right to left and calculate the count of smaller primes on the right of the current array element by getSum(). Print the count obtained.
• Now, check if the current array element is prime or not and update the Fenwick tree accordingly.

Below is the implementation of the above approach:

2 1 3 2 3 3 2 1 1 0 

Time Complexity: O(N log N)
Auxiliary Space: O(N)