This is a C++ program to count the inversion in an array.

1. The number of switches required to make an array sorted is termed as inversion count.

2. Its value varies with the sorting algorithms.

3. The time complexity is O(n^2).

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.

4. Exit.

C++ program to count the inversion in an array.

This program is successfully run on Dev-C++ using TDM-GCC 4.9.2 MinGW compiler on a Windows system.

#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 element: "; cin>>n; int arr[n]; for(i = 0; i < n; i++) { cout<<"Enter element "<<i+1<<": "; cin>>arr[i]; } // Printing the number of inversion in the array. cout<<"\nThe number of inversion in the array: "<<CountInversion(arr, n); return 0; }

1. Take input of data.

2. Call CountInversion() function with ‘arr’ the array of data and ‘n’ the number of values, in the argument list.

3. Using nested loops compare all the element with each other.

4. Increment the counter if a[i] > a[j] where i < j.
5. Return to main and display the result.
6. Exit.

Case 1: Enter the number of data element: 10 Enter element 1: 9 Enter element 2: 3 Enter element 3: 2 Enter element 4: 6 Enter element 5: 8 Enter element 6: 4 Enter element 7: 5 Enter element 8: 7 Enter element 9: 0 Enter element 10: 1 The number of inversion in the array: 29

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