Given an array arr[0..n-1]. The following operations need to be performed.

**update(l, r, val)**: Add ‘val’ to all the elements in the array from [l, r].**getRangeSum(l, r)**: Find sum of all elements in array from [l, r].

Initially all the elements in the array are 0. Queries can be in any oder, i.e., there can be many updates before range sum.

**Example:
**

Input : n = 5 // {0, 0, 0, 0, 0} Queries: update : l = 0, r = 4, val = 2 update : l = 3, r = 4, val = 3 getRangeSum : l = 2, r = 4 Output: Sum of elements of range [2, 4] is 12 Explanation : Array after first update becomes {2, 2, 2, 2, 2} Array after second update becomes {2, 2, 2, 5, 5}

In the previous post, we discussed range update and point query solutions using BIT.

rangeUpdate(l, r, val) : We add ‘val’ to element at index ‘l’. We subtract ‘val’ from element at index ‘r+1’.

getElement(index) [or getSum()]: We return sum of elements from 0 to index which can be quickly obtained using BIT.

We can compute rangeSum() using getSum() queries.

rangeSum(l, r) = getSum(r) – getSum(l-1)

A **Simple Solution **is to use solutions discussed in previous post. Range update query is same. Range sum query can be achieved by doing get query for all elements in range.

An** Efficient Solution **is to make sure that both queries can be done in O(Log n) time. We get range sum using prefix sums. How to make sure that update is done in a way so that prefix sum can be done quickly? Consider a situation where prefix sum [0, k] (where 0 <= k < n) is needed after range update on range [l, r]. Three cases arises as k can possibly lie in 3 regions.

**Case 1**: 0 < k < l

The update query won’t affect sum query.

**Case 2**: l <= k <= r

Consider an example:

Add 2 to range [2, 4], the resultant array would be: 0 0 2 2 2 If k = 3 Sum from [0, k] = 4

How to get this result?

Simply add the val from l^{th} index to k^{th} index. Sum is incremented by “val*(k) – val*(l-1)” after update query.

**Case 3**: k > r

For this case, we need to add “val” from l^{th} index to r^{th} index. Sum is incremented by “val*r – val*(l-1)” due to update query.

**Observations :**

**Case 1:** is simple as sum would remain same as it was before update.

**Case 2:** Sum was incremented by val*k – val*(l-1). We can find “val”, it is similar to finding the i^{th} element in range update and point query article. So we maintain one BIT for Range Update and Point Queries, this BIT will be helpful in finding the value at k^{th} index. Now val * k is computed, how to handle extra term val*(l-1)?

In order to handle this extra term, we maintain another BIT (BIT2). Update val * (l-1) at l^{th} index, so when getSum query is performed on BIT2 will give result as val*(l-1).

**Case 3 :** The sum in case 3 was incremented by “val*r – val *(l-1)”, the value of this term can be obtained using BIT2. Instead of adding, we subtract “val*(l-1) – val*r” as we can get this value from BIT2 by adding val*(l-1) as we did in case 2 and subtracting val*r in every update operation.

Update QueryUpdate(BITree1, l, val) Update(BITree1, r+1, -val) UpdateBIT2(BITree2, l, val*(l-1)) UpdateBIT2(BITree2, r+1, -val*r)Range SumgetSum(BITTree1, k) *k) - getSum(BITTree2, k)

C++ Implementation of above idea

// C++ program to demonstrate Range Update // and Range Queries using BIT #include <iostream> using namespace std; // Returns sum of arr[0..index]. This function assumes // that the array is preprocessed and partial sums of // array elements are stored in BITree[] int getSum(int BITree[], int index) { int sum = 0; // Initialize result // index in BITree[] is 1 more than the index in arr[] index = index + 1; // Traverse ancestors of BITree[index] while (index>0) { // Add current element of BITree to sum sum += BITree[index]; // Move index to parent node in getSum View index -= index & (-index); } return sum; } // Updates a node in Binary Index Tree (BITree) at given // index in BITree. The given value 'val' is added to // BITree[i] and all of its ancestors in tree. void updateBIT(int BITree[], int n, int index, int val) { // index in BITree[] is 1 more than the index in arr[] index = index + 1; // Traverse all ancestors and add 'val' while (index <= n) { // Add 'val' to current node of BI Tree BITree[index] += val; // Update index to that of parent in update View index += index & (-index); } } // Returns the sum of array from [0, x] int sum(int x, int BITTree1[], int BITTree2[]) { return (getSum(BITTree1, x) * x) - getSum(BITTree2, x); } void updateRange(int BITTree1[], int BITTree2[], int n, int val, int l, int r) { // Update Both the Binary Index Trees // As discussed in the article // Update BIT1 updateBIT(BITTree1,n,l,val); updateBIT(BITTree1,n,r+1,-val); // Update BIT2 updateBIT(BITTree2,n,l,val*(l-1)); updateBIT(BITTree2,n,r+1,-val*r); } int rangeSum(int l, int r, int BITTree1[], int BITTree2[]) { // Find sum from [0,r] then subtract sum // from [0,l-1] in order to find sum from // [l,r] return sum(r, BITTree1, BITTree2) - sum(l-1, BITTree1, BITTree2); } int *constructBITree(int n) { // Create and initialize BITree[] as 0 int *BITree = new int[n+1]; for (int i=1; i<=n; i++) BITree[i] = 0; return BITree; } // Driver Program to test above function int main() { int n = 5; // Construct two BIT int *BITTree1, *BITTree2; // BIT1 to get element at any index // in the array BITTree1 = constructBITree(n); // BIT 2 maintains the extra term // which needs to be subtracted BITTree2 = constructBITree(n); // Add 5 to all the elements from [0,4] int l = 0 , r = 4 , val = 5; updateRange(BITTree1,BITTree2,n,val,l,r); // Add 2 to all the elements from [2,4] l = 2 , r = 4 , val = 10; updateRange(BITTree1,BITTree2,n,val,l,r); // Find sum of all the elements from // [1,4] l = 1 , r = 4; cout << "Sum of elements from [" << l << "," << r << "] is "; cout << rangeSum(l,r,BITTree1,BITTree2) << "\n"; return 0; }

Output:

Sum of elements from [1,4] is 50

**Time Complexity** : O(q*log(n)) where q is number of queries.

This article is contributed by **Chirag Agarwal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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