Counting inversions in an array

Question

I\'m designing an algorithm to do the following: Given array A[1... n], for every i < j, find all inversion pairs such that A[i] > A[j]

Answer 1

Well I have a different solution but I am afraid that would work only for distinct array elements.

//Code
#include 
using namespace std;

int main()
{
    int i,n;
    cin >> n;
    int arr[n],inv[n];
    for(i=0;i> arr[i];
    }
    vector v;
    v.push_back(arr[n-1]);
    inv[n-1]=0;
    for(i=n-2;i>=0;i--){
        auto it = lower_bound(v.begin(),v.end(),arr[i]); 
        //calculating least element in vector v which is greater than arr[i]
        inv[i]=it-v.begin();
        //calculating distance from starting of vector
        v.insert(it,arr[i]);
        //inserting that element into vector v
    }
    for(i=0;i

To explain my code we keep on adding elements from the end of Array.For any incoming array element we find the index of first element in vector v which is greater than our incoming element and assign that value to inversion count of the index of incoming element.After that we insert that element into vector v at it's correct position such that vector v remain in sorted order.

//INPUT     
4
2 1 4 3

//OUTPUT    
1 0 1 0

//To calculate total inversion count just add up all the elements in output array