Sum of Interval and Update with Number of Divisors

Given an array A of N integers. You have to answer two types of queries :
1. Update [l, r] – for every i in range from l to r update Ai with D(Ai), where D(Ai) represents the number of divisors of Ai
2. Query [l, r] – calculate the sum of all numbers ranging between l and r in array A.
Input is given as two integers N and Q, representing number of integers in array and number of queries respectively. Next line contains an array of n integers followed by Q queries where ith query is represented as typei, li, ri.

Prerequisite : Binary Indexed Trees | Segment Trees

Examples :

Input : 7 4
        6 4 1 10 3 2 4
        2 1 7
        2 4 5
        1 3 5
        2 4 4
Output : 30
         13
         4

Explanation : First query is to calculate the sum of numbers from A1 to A7 which is 6 + 4
+ 1 + 10 + 3 + 2 + 4 = 30. Similarly, second query results into 13. For third query,
which is update operation, hence A3 will remain 1, A4 will become 4 and A5 will become 2.
Fourth query will result into A4 = 4.

Naive Approach :
A simple solution is to run a loop from l to r and calculate sum of elements in given range. To update a value, precompute the values of number of divisors of every number and simply do arr[i] = divisors[arr[i]].

Efficient Approach :
The idea is to reduce the time complexity for each query and update operation to O(logN). Use Binary Indexed Trees (BIT) or Segment Trees. Construct a BIT[] array and have two functions for query and update operation and precompute the number of divisors for each number. Now, for each update operation the key observation is that the numbers ‘1’ and ‘2’ will have ‘1’ and ‘2’ as their number of divisors respectively, so if it exists in the range of update query, they don’t need to be updated. We will use a set to store the index of only those numbers which are greater than 2 and use binary search to find the l index of the update query and increment the l index until every element is updated in range of that update query. If the arr[i] has only 2 divisors then after updating it, remove it from the set as it will always be 2 even after any next update query. For sum query operation, simply do query(r) – query(l – 1).
[sourcecode language=”CPP”]
// CPP program to calculate sum
// in an interval and update with
// number of divisors
#include <bits/stdc++.h>
using namespace std;

int divisors[100], BIT[100];

// structure for queries with members type,
// leftIndex, rightIndex of the query
struct queries
{
int type, l, r;
};

// function to calculate the number
// of divisors of each number
void calcDivisors()
{
for (int i = 1; i < 100; i++) {
for (int j = i; j < 100; j += i) {
divisors[j]++;
}
}
}

// function for updating the value
void update(int x, int val, int n)
{
for (x; x <= n; x += x&-x) {
BIT[x] += val;
}
}

// function for calculating the required
// sum between two indexes
int sum(int x)
{
int s = 0;
for (x; x > 0; x -= x&-x) {
s += BIT[x];
}
return s;
}

// function to return answer to queries
void answerQueries(int arr[], queries que[], int n, int q)
{
// Declaring a Set
set<int> s;
for (int i = 1; i < n; i++) {

// inserting indexes of those numbers
// which are greater than 2
if(arr[i] > 2) s.insert(i);
update(i, arr[i], n);
}

for (int i = 0; i < q; i++) {

// update query
if (que[i].type == 1) {
while (true) {

// find the left index of query in
// the set using binary search
auto it = s.lower_bound(que[i].l);

// if it crosses the right index of
// query or end of set, then break
if(it == s.end() || *it > que[i].r) break;

que[i].l = *it;

// update the value of arr[i] to
// its number of divisors
update(*it, divisors[arr[*it]] – arr[*it], n);

arr[*it] = divisors[arr[*it]];

// if updated value becomes less than or
// equal to 2 remove it from the set
if(arr[*it] <= 2) s.erase(*it);

// increment the index
que[i].l++;
}
}

// sum query
else {
cout << (sum(que[i].r) – sum(que[i].l – 1)) << endl;
}
}
}

// Driver Code
int main()
{
// precompute the number of divisors for each number
calcDivisors();

int q = 4;

// input array
int arr[] = {0, 6, 4, 1, 10, 3, 2, 4};
int n = sizeof(arr) / sizeof(arr[0]);

// declaring array of structure of type queries
queries que[q + 1];

que[0].type = 2, que[0].l = 1, que[0].r = 7;
que[1].type = 2, que[1].l = 4, que[1].r = 5;
que[2].type = 1, que[2].l = 3, que[2].r = 5;
que[3].type = 2, que[3].l = 4, que[3].r = 4;

// answer the Queries
answerQueries(arr, que, n, q);

return 0;
}
[/sourcecode]

Output:

30
13
4

Time Complexity for answering Q queries will be O(Q * log(N)).



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