### C++ program to count the inversions in the given array

Introduction
An inversion is formed between  two elements a[i] and a[j], if a[i] > a[j] and i<j.
Let us assume a array
3 5 6 9 1 2 7 8
Possible inversion in the array are:
Inversions: 10
Explanation: (3,1), (3,2), (5,1), (5,2), (6,1), (6,2)
(9,1), (9,2), (9,7), (9,8)
Algorithm:
1. Compare the values of the element with each other.
2. Increment the counter if the value at lower index is higher.
3. Display the result.

Program:
#include<iostream>

using namespace std;

int CountInversion(int a[], int n)
{
int i, j, count = 0;
for(i = 0; i < n; i++)
{
for(j = i+1; j < n; j++)
if(a[i] > a[j])
count++;
}
return count;
}

int main()
{
int n, i;
cout<<"\nEnter the number of data elements: "<<endl;
cin>>n;

int arr[n];
cout<<"Enter elements : ";
for(i = 0; i < n; i++)
{
cin>>arr[i];
}

// Printing the number of  inversion in the array.
cout<<"\nThe number of  inversion in the array: "<<CountInversion(arr, n);

return 0;
}
Output:
Enter the number of data elements:
8
Enter elements : 3 5 6 9 1 2 7 8

The number of  inversion in the array: 10

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