C++ Program to Count Inversion in an Array

Input / Output Scenario

Let's see the following input output scenario:

Input: arr[] = {6, 4, 2, 1}
Output: 6

Input: arr[] = {1, 2, 3, 4}
Output: 0

In the above scenario, there is no pair of indices (i, j) that exists in the given array such that arr[i]>arr[j] and i

Count Inversion in an Array Using Naive Approach

For each index in the array, we will count how many smaller elements are present on its right side. Then we will use a nested loop to calculate the inversion count and return the total.

Example

In the following example, we use the naive approach in C++ to count the inversions in the array -

#include 
#include 
using namespace std;

// Function to count inversions in the array
int inversionCount(vector < int > & arr) {
   int n = arr.size();
   int invCount = 0;

   // Loop through the array
   for (int i = 0; i < n - 1; i++) {
      for (int j = i + 1; j < n; j++) {
         if (arr[i] > arr[j])
            invCount++;
      }
   }
   return invCount;
}
int main() {
   vector < int > arr = {6, 4, 2, 1};
   cout << inversionCount(arr) << endl;
   return 0;
}

Following is the output of the above code -

6

Count Inversion in an Array Using Merge Sort

We can use merge sort to count the number of inversions in an array efficiently in O(n log n) time and O(n) space. First, we split the array into two halves: the left half and the right half. Then, we recursively count the number of inversions in each half. While merging the two halves back together, we also count the number of inversions that happen between the left and right halves (when an element from the left half is greater than one from the right half). Finally, we add up the inversions from the left, right, and between the halves to get the total number of inversions in the array.

Example

In the following example, we use the merge sort in C++ to count the inversions in the array -

#include 
#include 
using namespace std;

int countAndMerge(vector < int > & arr, int left, int mid, int right) {
   int n1 = mid - left + 1, n2 = right - mid;

   // left half vector
   vector < int > L(n1);
   // right half vector
   vector < int > R(n2);
   for (int i = 0; i < n1; i++)
      L[i] = arr[i + left];
   for (int j = 0; j < n2; j++)
      R[j] = arr[mid + 1 + j];

   // Initialize inversion count and merge two halves
   int res = 0;
   int i = 0, j = 0, k = left;
   while (i < n1 && j < n2) {
      if (L[i] <= R[j])
         arr[k++] = L[i++];
      else {
         arr[k++] = R[j++];
         res += (n1 - i);
      }
   }

   // Merge remaining elements
   while (i < n1)
      arr[k++] = L[i++];
   while (j < n2)
      arr[k++] = R[j++];

   return res;
}

// Function to count inversions in the array
int countInv(vector < int > & arr, int left, int right) {
   int res = 0;
   if (left < right) {
      int mid = (right + left) / 2;

      // Recursively count inversions in left and right halves
      res += countInv(arr, left, mid);
      res += countInv(arr, mid + 1, right);

      // Count split inversions
      res += countAndMerge(arr, left, mid, right);
   }
   return res;
}

int inversionCount(vector < int > & arr) {
   int n = arr.size();
   return countInv(arr, 0, n - 1);
}
int main() {
   vector < int > arr = {6, 4, 2, 1};
   cout << inversionCount(arr);
   return 0;
}

Following is the output of the code ?

6