Number of primes in a subarray (with updates)

Given an array of N integers, the task is to perform the following two queries:

query(start, end) : Print the number of prime numbers in the subarray from start to end 
update(i, x) : update the value at index i to x, i.e arr[i] = x

Examples:  

Input : arr = {1, 2, 3, 5, 7, 9}
        Query 1: query(start = 0, end = 4)
        Query 2: update(i = 3, x = 6)
        Query 3: query(start = 0, end = 4)
Output :4
        3
Explanation:
In Query 1, the subarray [0...4]
has 4 primes viz. {2, 3, 5, 7}

In Query 2, the value at index 3 
is updated to 6, the array arr now is, {1, 2, 3, 
6, 7, 9}
In Query 3, the subarray [0...4]
has 3 primes viz. {2, 3, 7}

Method 1 (Brute Force): A similar problem can be found here. Here there are no updates.We can modify this to handle updates but for this we need to build the prefix array always when we perform an update which makes the time complexity of this approach O(Q * N)

Method 2 (Efficient): Since, we need to handle both range queries and point updates, a segment tree is best suited for this purpose. 
We can use Sieve of Eratosthenes to preprocess all the primes till the maximum value arr_i can take say MAX in O(MAX log(log(MAX)).

Building the segment tree: We basically reduce the problem to subarray sum using segment tree.

Now, we can build the segment tree where a leaf node is represented as either 0 (if it is not a prime number) or 1 (if it is a prime number).
The internal nodes of the segment tree equal to the sum of its child nodes, thus a node represents the total primes in the range from L to R where the range L to R falls under this node and the sub-tree below it.

Handling Queries and Point Updates: Whenever we get a query from start to end, then we can query the segment tree for the sum of nodes in range start to end, which in turn represent the number of primes in range start to end. 

If we need to perform a point update and update the value at index i to x, then we check for the following cases: 

Let the old value of arri be y and the new value be x

Case 1: If x and y both are primes
Count of primes in the subarray does not change so we just update array and donot
modify the segment tree

Case 2: If x and y both are non primes
Count of primes in the subarray does not change so we just update array and donot
modify the segment tree

Case 3: If y is prime but x is non prime
Count of primes in the subarray decreases so we update array and add -1 to every 
range, the index i which is to be updated, is a part of in the segment tree

Case 4: If y is non prime but x is prime
Count of primes in the subarray increases so we update array and add 1 to every 
range, the index i which is to be updated, is a part of in the segment tree

Implementation:

Number of Primes in subarray from 0 to 4 = 4
Number of Primes in subarray from 0 to 4 = 3

The time complexity of each query and update is O(logn) and that of building the segment tree is O(n) 

Note: Here, the time complexity of pre-processing primes till MAX using the sieve of Eratosthenes is O(MAX log(log(MAX))) where MAX is the maximum value arr_i can take.

Auxiliary Space: O(MAX+n)

Related Topic: Segment Tree