**Prerequisites:** Segment Tree

Given a number **N** which represents the size of the array initialized to 0 and **Q** queries to process where there are two types of queries:

- 1 P V: Put the value
**V**at position**P**. - 2 L R: Output the sum of values from
**L**to**R**.

The task is to answer these queries.

**Constraints:**

- 1 ≤ N ≤ 10
^{18} - Q ≤ 10
^{5} - 1 ≤ L ≤ R≤ N

**Note:** Queries are online. Therefore:

- L = (previousAnswer + L) % N + 1
- R = (previousAnswer + R) % N + 1

**Examples:**

Input:N = 5, Q = 5, arr[][] = {{1, 2, 3}, {1, 1, 4}, {1, 3, 5}, {1, 4, 7}, {2, 3, 4}}

Output:12

Explanation:

There are five queries. Since N = 5, therefore, initially, the array is {0, 0, 0, 0, 0}

For query 1: 1 2 3 array = {0, 3, 0, 0, 0}

For query 2: 1 1 4 array = {4, 3, 0, 0, 0}

For query 3: 1 3 5 array = {4, 3, 5, 0, 0}

For query 4: 1 4 7 array = {4, 3, 5, 7, 0}

For query 5: 2 3 4 Sum from [3, 4] = 7 + 5 = 12.

Input:N = 3, Q = 2, arr[][] = {{1, 1, 1}, {1, 2, 2}, {1, 3, 3}}

Output:0

**Approach:** Here, since the updates are high, Kadane’s algorithm doesn’t work quite well. Moreover, since it is given that the queries are online, a simple segment tree would not be able to solve this problem because the constraints for the number of elements is very high. Therefore, a new type of data structure, a dynamic segment tree is used in this problem.

**Dynamic Segment Tree:** Dynamic segment tree is not a new data structure. It is very similar to the segment tree. The following are the properties of the dynamic segment tree:

- Instead of using an array to represent the intervals, a node is created whenever a new interval is to be updated.
- The following is the structure of the node of the dynamic segment tree:
- Clearly, the above structure is the same as a Binary Search Tree. In every node, we are storing the node’s value and two pointers pointing to the left and right subtree.
- The Interval of the root is from [1, N], the interval of the left subtree will be [1, N/2] and the interval for the right subtree will be [N/2 + 1, N].
- Similarly, for every node, we can calculate the interval it is representing. Let’s say the interval of the current node is [L, R]. Then, the Interval of its left and right subtree are [L, (L + R)/2] and [(L + R)/2+1, R] respectively.
- Since we are creating a new node only when required, the build() function from the segment tree is completely removed.

`// Every node contains the value and ` `// the left subtree and right subtree ` `struct` `Node { ` ` ` `long` `long` `value; ` ` ` `struct` `Node *L, *R; ` `}; ` ` ` `struct` `Node* getnode() ` `{ ` ` ` `struct` `Node* temp = ` `new` `struct` `Node; ` ` ` `temp->value = 0; ` ` ` `temp->L = NULL; ` ` ` `temp->R = NULL; ` ` ` `return` `temp; ` `} ` |

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Before getting into the algorithm for the operations, let’s define the terms used in this article:

**Node’s interval:**It is the interval the node is representing.**Required interval:**Interval for which the sum is to calculate.**Required index:**Index at which Update is required.

The following is the algorithm used for the operations on the tree with the above-mentioned properties:

**Point Update:**The following algorithm is used for the point update:- Start with the root node.
- If the interval at the node doesn’t overlap with the required index then return.
- If the node is a NULL entry then create a new node with the appropriate intervals and descend into that node by going back to step 2 for every new child created.
- If both, intervals and the index at which the value is to be stored are equal, then store the value into at that node.
- If the interval at the node overlaps partially with the required index then descend into its children and continue the execution from step 2.

**Finding the sum for every query:**The following algorithm is used to find the sum for every query:- Start with the root node.
- If the node is a NULL or the interval at that node doesn’t overlap with the required interval, then return 0.
- If the interval at the node completely overlaps with the required interval then return the value stored at the node.
- If the interval at the node overlaps partially with the required interval then descend into its children and continue the execution from step 2 for both of its children.

**Example:** Lets visualize the update and sum with an example. Let N = 10 and the operations needed to perform on the tree are as follows:

- Insert 10 at position 1.
- Find the sum of value of indices from 2 to 8.
- Insert 3 at position 5.
- Find the sum of value of indices from 3 to 6.

- Initially, for the value N = 10, the tree is empty. Therefore:

- Insert 10 at position 1. In order to do this, create a new node until we get the required interval. Therefore:

- Find the sum of value of indices from 2 to 8. In order to do this, the sum from [1, 8] is found and the value [1, 2] is subtracted from it. Since the node [1, 8] is not yet created, the value of [1, 8] is the value of the root [1, 10]. Therefore:

- Insert 3 at position 5. In order to do this, create a new node until we get the required interval. Therefore:

Below is the implementation of the above approach:

`// C++ program for the implementation ` `// of the Dynamic segment tree and ` `// perform the range updates on the ` `// given queries ` ` ` `#include <bits/stdc++.h> ` ` ` `using` `namespace` `std; ` `typedef` `long` `long` `ll; ` ` ` `// Structure of the node ` `struct` `Node { ` ` ` ` ` `ll value; ` ` ` `struct` `Node *L, *R; ` `}; ` ` ` `// Structure to get the newly formed ` `// node ` `struct` `Node* getnode() ` `{ ` ` ` `struct` `Node* temp = ` `new` `struct` `Node; ` ` ` `temp->value = 0; ` ` ` `temp->L = NULL; ` ` ` `temp->R = NULL; ` ` ` `return` `temp; ` `} ` ` ` `// Creating the Root node ` `struct` `Node* root; ` ` ` `// Function to perform the point update ` `// on the dynamic segment tree ` `void` `UpdateHelper(` `struct` `Node* curr, ll index, ` ` ` `ll L, ll R, ll val) ` `{ ` ` ` ` ` `// If the index is not overlapping ` ` ` `// with the index ` ` ` `if` `(L > index || R < index) ` ` ` `return` `; ` ` ` ` ` `// If the index is completely overlapping ` ` ` `// with the index ` ` ` `if` `(L == R && L == index) { ` ` ` ` ` `// Update the value of the node ` ` ` `// to the given value ` ` ` `curr->value = val; ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `// Computing the middle index if none ` ` ` `// of the above base cases are satisfied ` ` ` `ll mid = L - (L - R) / 2; ` ` ` `ll sum1 = 0, sum2 = 0; ` ` ` ` ` `// If the index is in the left subtree ` ` ` `if` `(index <= mid) { ` ` ` ` ` `// Create a new node if the left ` ` ` `// subtree is is null ` ` ` `if` `(curr->L == NULL) ` ` ` `curr->L = getnode(); ` ` ` ` ` `// Recursively call the function ` ` ` `// for the left subtree ` ` ` `UpdateHelper(curr->L, index, L, mid, val); ` ` ` `} ` ` ` ` ` `// If the index is in the right subtree ` ` ` `else` `{ ` ` ` ` ` `// Create a new node if the right ` ` ` `// subtree is is null ` ` ` `if` `(curr->R == NULL) ` ` ` `curr->R = getnode(); ` ` ` ` ` `// Recursively call the function ` ` ` `// for the right subtree ` ` ` `UpdateHelper(curr->R, index, mid + 1, R, val); ` ` ` `} ` ` ` ` ` `// Storing the sum of the left subtree ` ` ` `if` `(curr->L) ` ` ` `sum1 = curr->L->value; ` ` ` ` ` `// Storing the sum of the right subtree ` ` ` `if` `(curr->R) ` ` ` `sum2 = curr->R->value; ` ` ` ` ` `// Storing the sum of the children into ` ` ` `// the node's value ` ` ` `curr->value = sum1 + sum2; ` ` ` `return` `; ` `} ` ` ` `// Function to find the sum of the ` `// values given by the range ` `ll queryHelper(` `struct` `Node* curr, ll a, ` ` ` `ll b, ll L, ll R) ` `{ ` ` ` ` ` `// Return 0 if the root is null ` ` ` `if` `(curr == NULL) ` ` ` `return` `0; ` ` ` ` ` `// If the index is not overlapping ` ` ` `// with the index, then the node ` ` ` `// is not created. So sum is 0 ` ` ` `if` `(L > b || R < a) ` ` ` `return` `0; ` ` ` ` ` `// If the index is completely overlapping ` ` ` `// with the index, return the node's value ` ` ` `if` `(L >= a && R <= b) ` ` ` `return` `curr->value; ` ` ` ` ` `ll mid = L - (L - R) / 2; ` ` ` ` ` `// Return the sum of values stored ` ` ` `// at the node's children ` ` ` `return` `queryHelper(curr->L, a, b, L, mid) ` ` ` `+ queryHelper(curr->R, a, b, mid + 1, R); ` `} ` ` ` `// Function to call the queryHelper ` `// function to find the sum for ` `// the query ` `ll query(` `int` `L, ` `int` `R) ` `{ ` ` ` `return` `queryHelper(root, L, R, 1, 10); ` `} ` ` ` `// Function to call the UpdateHelper ` `// function for the point update ` `void` `update(` `int` `index, ` `int` `value) ` `{ ` ` ` `UpdateHelper(root, index, 1, 10, value); ` `} ` ` ` `// Function to perform the operations ` `// on the tree ` `void` `operations() ` `{ ` ` ` `// Creating an empty tree ` ` ` `root = getnode(); ` ` ` ` ` `// Update the value at position 1 to 10 ` ` ` `update(1, 10); ` ` ` ` ` `// Update the value at position 3 to 5 ` ` ` `update(3, 5); ` ` ` ` ` `// Finding sum for the range [2, 8] ` ` ` `cout << query(2, 8) << endl; ` ` ` ` ` `// Finding sum for the range [1, 10] ` ` ` `cout << query(1, 10) << endl; ` ` ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `operations(); ` ` ` ` ` `return` `0; ` `} ` |

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**Output:**

5 15

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- Two equal sum segment range queries
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- Segment Trees | (Product of given Range Modulo m)
- Queries for elements greater than K in the given index range using Segment Tree
- Queries for the count of even digit sum elements in the given range using Segment Tree.
- Queries for greatest pair sum in the given index range using Segment Tree
- Count of distinct numbers in an Array in a range for Online Queries using Merge Sort Tree

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