Advanced Applications of Fenwick Trees

Fenwick Trees, also known as Binary Indexed Trees (BIT), have a wide range of applications beyond just computing prefix sums and performing range queries.

One of the advanced applications of Fenwick Trees is solving the problem of finding the k-th element in a sorted array. Given an initial array, the Fenwick Tree can help efficiently perform insertions and deletions, while still being able to find the k-th element in O(log N) time complexity.

Another application of Fenwick Trees is solving the problem of finding the median of a set of numbers. By maintaining two Fenwick Trees, one for the left half of the numbers and the other for the right half, we can easily find the median in O(log N) time complexity.

Furthermore, Fenwick Trees can also be used to solve the problem of finding the inversion count in an array. An inversion in an array occurs when two elements are out of order. By using a Fenwick Tree, we can efficiently count the number of inversions in an array in O(N log N) time complexity.

Let's take a look at the implementation of finding the inversion count in an array using Fenwick Trees in C++:

1#include <iostream>
2using namespace std;
3
4int getSum(int BIT[], int index) {
5  int sum = 0;
6  while (index > 0) {
7    sum += BIT[index];
8    index -= index & (-index);
9  }
10  return sum;
11}
12
13void updateBIT(int BIT[], int n, int index, int value) {
14  while (index <= n) {
15    BIT[index] += value;
16    index += index & (-index);
17  }
18}
19
20int getInversionCount(int arr[], int n) {
21  int inversionCount = 0;
22  int maxElement = 0;
23
24  for (int i = 0; i < n; i++) {
25    if (arr[i] > maxElement) {
26      maxElement = arr[i];
27    }
28  }
29
30  int BIT[maxElement + 1];
31  for (int i = 1; i <= maxElement; i++) {
32    BIT[i] = 0;
33  }
34
35  for (int i = n - 1; i >= 0; i--) {
36    inversionCount += getSum(BIT, arr[i] - 1);
37    updateBIT(BIT, maxElement, arr[i], 1);
38  }
39
40  return inversionCount;
41}
42
43int main() {
44  int arr[] = {8, 4, 2, 1};
45  int n = sizeof(arr) / sizeof(arr[0]);
46
47  int inversionCount = getInversionCount(arr, n);
48
49  cout << "Inversion Count: " << inversionCount << endl;
50
51  return 0;
52}

In the above example, we initialize a Fenwick Tree of size maxElement + 1 to maintain the count of each element occurrence in the array. Then, we iterate over the array from right to left and count the number of elements smaller than the current element on the right side. This gives us the inversion count of the array.

By exploring these advanced applications, you'll be able to leverage the power of Fenwick Trees in a variety of problem-solving scenarios.

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#include <iostream>

using namespace std;

​

int main() {

  // Replace with your C++ logic here

​

  return 0;

}