Given an array arr[] of size n. Three elements arr[i], arr[j] and arr[k] form an inversion of size 3 if a[i] > a[j] >a[k] and i < j < k. Find total number of inversions of size 3.
Example :
Input: {8, 4, 2, 1}
Output: 4
The four inversions are (8,4,2), (8,4,1), (4,2,1) and (8,2,1).
Input: {9, 6, 4, 5, 8}
Output: 2
The two inversions are {9, 6, 4} and {9, 6, 5}
We have already discussed inversion count of size two by merge sort, Self Balancing BST and BIT.
Simple approach :- Loop for all possible value of i, j and k and check for the condition a[i] > a[j] > a[k] and i < j < k.
Java
// A simple Java implementation to count inversion of size 3class Inversion{ // returns count of inversion of size 3 int getInvCount(int arr[], int n) { int invcount = 0; // initialize result for(int i=0 ; i< n-2; i++) { for(int j=i+1; j<n-1; j++) { if(arr[i] > arr[j]) { for(int k=j+1; k<n; k++) { if(arr[j] > arr[k]) invcount++; } } } } return invcount; } // driver program to test above function public static void main(String args[]) { Inversion inversion = new Inversion(); int arr[] = new int[] {8, 4, 2, 1}; int n = arr.length; System.out.print("Inversion count : " + inversion.getInvCount(arr, n)); }}// This code is contributed by Mayank Jaiswal |
Output:
Inversion Count : 4
Time complexity of this approach is : O(n^3)
Better Approach :
We can reduce the complexity if we consider every element arr[i] as middle element of inversion, find all the numbers greater than a[i] whose index is less than i, find all the numbers which are smaller than a[i] and index is more than i. We multiply the number of elements greater than a[i] to the number of elements smaller than a[i] and add it to the result.
Below is the implementation of the idea.
Java
// A O(n^2) Java program to count inversions of size 3 class Inversion { // returns count of inversion of size 3 int getInvCount(int arr[], int n) { int invcount = 0; // initialize result for (int i=0 ; i< n-1; i++) { // count all smaller elements on right of arr[i] int small=0; for (int j=i+1; j<n; j++) if (arr[i] > arr[j]) small++; // count all greater elements on left of arr[i] int great = 0; for (int j=i-1; j>=0; j--) if (arr[i] < arr[j]) great++; // update inversion count by adding all inversions // that have arr[i] as middle of three elements invcount += great*small; } return invcount; } // driver program to test above function public static void main(String args[]) { Inversion inversion = new Inversion(); int arr[] = new int[] {8, 4, 2, 1}; int n = arr.length; System.out.print("Inversion count : " + inversion.getInvCount(arr, n)); }} // This code has been contributed by Mayank Jaiswal |
Output :
Inversion Count : 4
Time Complexity of this approach : O(n^2)
Binary Indexed Tree Approach :
Like inversions of size 2, we can use Binary indexed tree to find inversions of size 3. It is strongly recommended to refer below article first.
Count inversions of size two Using BIT
The idea is similar to above method. We count the number of greater elements and smaller elements for all the elements and then multiply greater[] to smaller[] and add it to the result.
Solution :
- To find out the number of smaller elements for an index we iterate from n-1 to 0. For every element a[i] we calculate the getSum() function for (a[i]-1) which gives the number of elements till a[i]-1.
- To find out the number of greater elements for an index we iterate from 0 to n-1. For every element a[i] we calculate the sum of numbers till a[i] (sum smaller or equal to a[i]) by getSum() and subtract it from i (as i is the total number of element till that point) so that we can get number of elements greater than a[i].
Please refer complete article on Count Inversions of size three in a given array for more details!
