Queries for Composite numbers in subarray (With Point Updates)

Given an array of N integers, the task is to perform the following two operations on the given array:

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


Input : arr = {1, 12, 3, 5, 17, 9}
        Query 1: query(start = 0, end = 4)
        Query 2: update(i = 3, x = 6)
        Query 3: query(start = 0, end = 4)
Output :1
In Query 1, the subarray [0...4]
has 1 Composite number viz. {12}

In Query 2, the value at index 3 
is updated to 6, the array arr now is, {1, 12, 3, 
6, 7, 9}
In Query 3, the subarray [0...4]
has 2 Composite Numbers viz. {12, 6}

Since, we need to handle both range queries and point updates, an efficient method is to use a segment tree to solve the problem. A segment tree is best suited for this purpose.

We can use Sieve of Eratosthenes to preprocess all the primes till the maximum value that arri can take, say MAX. The time complexity for this operation will be O(MAX log(log(MAX))).

Building the segment tree:
The problem can be reduced 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 a prime number) or 1 (if it is a composite number).

The internal nodes of the segment tree equal to the sum of its child nodes, thus a node represents the total composite numbers 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 composites 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 composites.
Count of composites 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 primes.
Count of composites in the subarray does not change so we just update array and donot
modify the segment tree

Case 3: If y is composite but x is prime.
Count of composite numbers 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 prime but x is composite.
Count of composite numbers 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

Below is the implementation of the above approach:





// C++ program to find number of composite numbers in a
// subarray and performing updates
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000
// Function to calculate primes upto MAX
// using sieve of Eratosthenes
void sieveOfEratosthenes(bool isPrime[])
    isPrime[1] = true;
    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 composites 
    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 queryCompositesUtil(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 composites
    // 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 queryCompositesUtil(st, ss, mid, qs, qe, 2 * index + 1)
           + queryCompositesUtil(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)
    // 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");
    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])
    // Case 2: Old and new values both composite
    if ((!isPrime[oldValue]) && (!isPrime[new_val]))
    // Case 3: Old value was composite, new value is prime
    if (!isPrime[oldValue] && isPrime[new_val]) {
        diff = -1;
    // Case 4: Old value was prime, new_val is composite
    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 composite numbers in range
// from index qs (query start) to qe (query end).
// It mainly uses queryCompositesUtil()
void queryComposites(int* st, int n, int qs, int qe)
    int compositesInRange = queryCompositesUtil(st, 0, n - 1, qs, qe, 0);
    cout << "Number of Composites in subarray from " << qs
         << " to " << qe << " = " << compositesInRange << "\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 composite
        if (!isPrime[arr[ss]])
            st[si] = 1;
            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 Code
int main()
    int arr[] = { 1, 12, 3, 5, 17, 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 composite, else true. 
    bool isPrime[MAX + 1];
    memset(isPrime, true, sizeof 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;
    queryComposites(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);
    // Query 3: Query(start = 0, end = 4)
    start = 0;
    end = 4;
    queryComposites(st, n, start, end);
    return 0;



Number of Composites in subarray from 0 to 4 = 1
Number of Composites in subarray from 0 to 4 = 2

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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