Let’s talk about Fenwick trees(a.k.a. Binary Indexed Trees) for a moment and see how we can use them to solve range problems. This is not a tutorial, just a few notes for my future self. By the way, I already posted about this topic some time ago, this time, however, I’ll focus my attention on a particular kind of problem. Also, more practice problems.

Range problems usually have this format: given a list of values, we’d like to support the following operations:

- Update value at index i
- Compute f(a, b)

Or

- Apply operation X to all elements in the range [a, b]
- Get the value at index i

Whenever you see a problem like this you should think about Fenwick trees and there is a good chance it will be the data structure you need to solve the problem.

## Problem & solution

Let’s take a look at the following problem from LeetCode Range Sum Query – Mutable.

If you know about Fenwick trees you’ll recognize the problem immediately and hopefully come up with a solution quickly. The problem follows the first pattern presented above. This is my solution:

class NumArray { private: vector<int> T; int lowestOneBit(int n) { return n & (-n); } void update(vector<int>& T, int index, int delta) { while (index < int(T.size())) { T[index] += delta; index += lowestOneBit(index); } } int read(vector<int>& T, int index) { int sum = 0; while (index > 0) { sum += T[index]; index -= lowestOneBit(index); } return sum; } public: NumArray(vector<int> nums) { T.resize(nums.size() + 1); // Index the initial values for (size_t index = 0; index < nums.size(); index++) { update(T, index + 1, nums[index]); } } void update(int i, int newValue) { // Compute the current value at index i int value = read(T, i + 1) - read(T, i); // Update the BIT with the difference, not the new value update(T, i + 1, newValue - value); } int sumRange(int i, int j) { return read(T, j + 1) - read(T, i); } };

## Practice problems

Here you have some additional practice problems to hone your skills, enjoy!

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