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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