How to find inversion pairs in an array and their index positions? - c++

Question

I find difficult to find the inversion pairs in the array. I got the number of inversion pairs in the array but the logic seems so complex that I can't able to list pairs. I am new to C++, any help is appreciated.
Example:
Input: arr[] = {8, 4, 2, 1}
Output: 6
Expecting this output
Given array has six inversions (8,4) and index(0,1), (4,2) and index(1,2),
(8,2) and index(0,2), (8,1) and index(0,3), (4,1) and index(1,3), (2,1) and index(2,4)
// C++ program to count inversions using Binary Indexed Tree
#include<bits/stdc++.h>
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
// 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)
{
// 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 inversion count arr[0..n-1]
int getInvCount(int arr[], int n)
{
int invcount = 0; // Initialize result
// Find maximum element in arr[]
int maxElement = 0;
for (int i=0; i<n; i++)
if (maxElement < arr[i])
maxElement = arr[i];
// Create a BIT with size equal to maxElement+1 (Extra
// one is used so that elements can be directly be
// used as index)
int BIT[maxElement+1];
for (int i=1; i<=maxElement; i++)
BIT[i] = 0;
// Traverse all elements from right.
for (int i=n-1; i>=0; i--)
{
// Get count of elements smaller than arr[i]
invcount += getSum(BIT, arr[i]-1);
// Add current element to BIT
updateBIT(BIT, maxElement, arr[i], 1);
}
return invcount;
}
// Driver program
int main()
{
int arr[] = {8, 4, 2, 1};
int n = sizeof(arr)/sizeof(int);
cout << "Number of inversions are : " << getInvCount(arr,n);
return 0;
}
The program creates new array BITree[] that makes it more complex. How the position of the element is found.