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 4 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 arri 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
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// C++ program to find number of prime numbers in a 
// subarray and performing updates
#include <bits/stdc++.h>
using namespace std;
  
#define MAX 1000
  
void sieveOfEratosthenes(bool isPrime[])
{
    isPrime[1] = false;
  
    for (int p = 2; p * p <= MAX; p++) {
  
        // If prime[p] is not changed, then
        // it is a prime
        if (isPrime[p] == true) {
  
            // Update all multiples of p
            for (int i = p * 2; i <= MAX; i += p)
                isPrime[i] = false;
        }
    }
}
  
// A utility function to get the middle index from corner indexes.
int getMid(int s, int e) { return s + (e - s) / 2; }
  
/*  A recursive function to get the number of primes in a given range
     of array indexes. The following are parameters for this function.
   
    st    --> Pointer to segment tree
    index --> Index of current node in the segment tree. Initially
              0 is passed as root is always at index 0
    ss & se  --> Starting and ending indexes of the segment represented
                  by current node, i.e., st[index]
    qs & qe  --> Starting and ending indexes of query range */
int queryPrimesUtil(int* st, int ss, int se, int qs, int qe, int index)
{
    // If segment of this node is a part of given range, then return
    // the number of primes in the segment
    if (qs <= ss && qe >= se)
        return st[index];
  
    // If segment of this node is outside the given range
    if (se < qs || ss > qe)
        return 0;
  
    // If a part of this segment overlaps with the given range
    int mid = getMid(ss, se);
    return queryPrimesUtil(st, ss, mid, qs, qe, 2 * index + 1) + 
           queryPrimesUtil(st, mid + 1, se, qs, qe, 2 * index + 2);
}
  
/* A recursive function to update the nodes which have the given 
   index in their range. The following are parameters
    st, si, ss and se are same as getSumUtil()
    i    --> index of the element to be updated. This index is 
             in input array.
   diff --> Value to be added to all nodes which have i in range */
void updateValueUtil(int* st, int ss, int se, int i, int diff, int si)
{
    // Base Case: If the input index lies outside the range of
    // this segment
    if (i < ss || i > se)
        return;
  
    // If the input index is in range of this node, then update
    // the value of the node and its children
    st[si] = st[si] + diff;
    if (se != ss) {
        int mid = getMid(ss, se);
        updateValueUtil(st, ss, mid, i, diff, 2 * si + 1);
        updateValueUtil(st, mid + 1, se, i, diff, 2 * si + 2);
    }
}
  
// The function to update a value in input array and segment tree.
// It uses updateValueUtil() to update the value in segment tree
void updateValue(int arr[], int* st, int n, int i, int new_val,
                                               bool isPrime[])
{
    // Check for erroneous input index
    if (i < 0 || i > n - 1) {
        printf("Invalid Input");
        return;
    }
  
    int diff, oldValue;
  
    oldValue = arr[i];
  
    // Update the value in array
    arr[i] = new_val;
  
    // Case 1: Old and new values both are primes
    if (isPrime[oldValue] && isPrime[new_val])
        return;
  
    // Case 2: Old and new values both non primes
    if ((!isPrime[oldValue]) && (!isPrime[new_val]))
        return;
  
    // Case 3: Old value was prime, new value is non prime
    if (isPrime[oldValue] && !isPrime[new_val]) {
        diff = -1;
    }
  
    // Case 4: Old value was non prime, new_val is prime
    if (!isPrime[oldValue] && isPrime[new_val]) {
        diff = 1;
    }
  
    // Update the values of nodes in segment tree
    updateValueUtil(st, 0, n - 1, i, diff, 0);
}
  
// Return number of primes in range from index qs (query start) to
// qe (query end).  It mainly uses queryPrimesUtil()
void queryPrimes(int* st, int n, int qs, int qe)
{
    int primesInRange = queryPrimesUtil(st, 0, n - 1, qs, qe, 0);
  
    cout << "Number of Primes in subarray from " << qs << " to "
         << qe << " = " << primesInRange << "\n";
}
  
// A recursive function that constructs Segment Tree 
// for array[ss..se].
// si is index of current node in segment tree st
int constructSTUtil(int arr[], int ss, int se, int* st, 
                                 int si, bool isPrime[])
{
    // If there is one element in array, check if it
    // is prime then store 1 in the segment tree else
    // store 0 and return
    if (ss == se) {
  
        // if arr[ss] is prime
        if (isPrime[arr[ss]]) 
            st[si] = 1;        
        else 
            st[si] = 0;
          
        return st[si];
    }
  
    // If there are more than one elements, then recur 
    // for left and right subtrees and store the sum 
    // of the two values in this node
    int mid = getMid(ss, se);
    st[si] = constructSTUtil(arr, ss, mid, st, 
                               si * 2 + 1, isPrime) + 
             constructSTUtil(arr, mid + 1, se, st, 
                              si * 2 + 2, isPrime);
    return st[si];
}
  
/* Function to construct segment tree from given array. 
   This function allocates memory for segment tree and
   calls constructSTUtil() to fill the allocated memory */
int* constructST(int arr[], int n, bool isPrime[])
{
    // Allocate memory for segment tree
  
    // Height of segment tree
    int x = (int)(ceil(log2(n)));
  
    // Maximum size of segment tree
    int max_size = 2 * (int)pow(2, x) - 1;
  
    int* st = new int[max_size];
  
    // Fill the allocated memory st
    constructSTUtil(arr, 0, n - 1, st, 0, isPrime);
  
    // Return the constructed segment tree
    return st;
}
  
// Driver program to test above functions
int main()
{
  
    int arr[] = { 1, 2, 3, 5, 7, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    /* Preprocess all primes till MAX.
       Create a boolean array "isPrime[0..MAX]".
       A value in prime[i] will finally be false 
       if i is Not a prime, else true. */
  
    bool isPrime[MAX + 1];
    memset(isPrime, true, sizeof isPrime);
    sieveOfEratosthenes(isPrime);
  
    // Build segment tree from given array
    int* st = constructST(arr, n, isPrime);
  
    // Query 1: Query(start = 0, end = 4)
    int start = 0;
    int end = 4;
    queryPrimes(st, n, start, end);
  
    // Query 2: Update(i = 3, x = 6), i.e Update 
    // a[i] to x
    int i = 3;
    int x = 6;
    updateValue(arr, st, n, i, x, isPrime);
  
    // uncomment to see array after update
    // for(int i = 0; i < n; i++) cout << arr[i] << " ";
  
    // Query 3: Query(start = 0, end = 4)
    start = 0;
    end = 4;
    queryPrimes(st, n, start, end);
  
    return 0;
}

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Output:

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 arri can take



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