# Java Program to count inversions of size three in a given array

Inversion count is a step counting method by which we can calculate the number of sorting steps taken by a particular array. It is also capable to count the operation time span for an array. But, if we want to sort an array in a reverse manner, the count will be maximum number present in that array.

Array: { 5, 4, 3, 2, 1}  // for the reverse manner
Pairs: {5, 4}, {5,3} , {3,2}, {3,1}, {2,1},{4,3}, {4,2}, {4,1},}, {5,2}, {5,1}
Output: 10
Array: {1, 2, 3, 4, 5}  // for the increasing manner
Pairs: No Pairs
Output: 0
Array: {1,5,2,8,3,4}
Pairs: {5, 2}, {5, 3}, {5, 4}, {8, 3}, {8, 4}
Output: 5


The inversion count indicates that how far that particular array is from being sorted in an increasing order. Here are two particular process to describe this situation attached with a solution −

• To find the smaller elements − To find out the smaller element from an array, we need to iterate the index from n-1 to 0. By applying (a[i]-1), we can calculate the getSum() here. The process will run until it reach to a[i]-1.

• To find the greater number − To find the greater number from an index we need to perform iteration 0 to n-1. For the every element we need to do calculation for every number till a[i]. Subtract it from i. Then we will get a the number which is greater than a[i].

### Algorithm to count inversions of size three in an array:-

Here in this algorithm; we learn how to count inversions of size three in a given array in a particular programming environment.

• Step 1 − Start

• Step 2 − Declare an array and inversion count (As arr[] --> array and invCount --> Inversion count)

• Step 3 − Inner loop y=x+1 to N

• Step 4 − If element at x is greater than element at y index

• Step 5 − Then, increase the invCount++

• Step 6 − Print the pair

• Step 7 − Terminate

### Syntax to count inversions of size three in an array:-

A pair (A[i], A[j]) is said to be in inversion if: A[i] > A[j] and i < j

C++ Implementation

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


Java Implementation

public static int getInversions(int[] A, int n) {
int count = 0;

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

}
}
return count;
}


Python Implementation

def getInversions(A, n):
count = 0
for i in range(n):
for j in range(i + 1, n):
if A[i] > A[j]:
count += 1

return count;
}


Here we have mentioned the possible syntaxes to count inversions of size three in a given array. And for this method; Time Complexity is O(N^2), where N is the total size of the array and; Space Complexity:O(1), as no extra space has been used.

### Approaches to follow

• Approach 1 − Count Inversions of size three in a given array by program to count inversions of size 3

• Approach 2 − Better Approach to count inversions of size 3

• Approach 3 − Count inversions of size 3 using binary indexed tree

## Count Inversions of size three in a given array by program to count inversions of size 3

For the simple approach to count inversions of size three, we need to run a loop for all possible value of i, j and k. The time complexity is O(n^3) and O(1) reflects the auxiliary space.

The condition is:

a[i] > a[j] > a[k] and i < j < k.

### Example 1

public class Inversion{
int getInvCount(int arr[], int n){
int invcount = 0;

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;
}
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 after method: " +
inversion.getInvCount(arr, n));
}
}


### Output

Inversion count after method: 4


## A better approach to count inversions of size 3

In this method we will consider the every element of an array as middle element of inversion. It helps to reduce the complexity. For this approach, the time complexity is O(n^2) and auxiliary Space is O(1).

### Example 2

public class Inversion {
int getInvCount(int arr[], int n){
int invcount = 0;

for (int i=0 ; i< n-1; i++){
int small=0;
for (int j=i+1; j<n; j++)
if (arr[i] > arr[j])
small++;
int great = 0;
for (int j=i-1; j>=0; j--)
if (arr[i] < arr[j])
great++;
invcount += great*small;
}
return invcount;
}
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 afret the operation : " +
inversion.getInvCount(arr, n));
}
}


### Output

Inversion count afret the operation : 4


## Count inversions of size 3 using binary indexed tree

In this method, we count the greater elements and smaller ones too. Then perform the multiply operation greater[] to smaller[] and add it to the final result. Here the time complexity is O(n*log(n)) and auxiliary space denoted as O(n).

### Example 3

import java.io.*;
import java.util.Arrays;
import java.util.ArrayList;
import java.lang.*;
import java.util.Collections;

public class rudrabytp {
static int N = 100005;
static int BIT[][] = new int[N];
static void updateBIT(int t, int i, int val, int n){
while (i <= n) {
BIT[t][i] = BIT[t][i] + val;
i = i + (i & (-i));
}
}
static int getSum(int t, int i){
int res = 0;
while (i > 0) {
res = res + BIT[t][i];
i = i - (i & (-i));
}
return res;
}
static void convert(int arr[], int n){
int temp[]=new int[n];
for (int i = 0; i < n; i++)
temp[i] = arr[i];
Arrays.sort(temp);
for (int i = 0; i < n; i++) {
arr[i] = Arrays.binarySearch(temp,arr[i]) + 1;
}
}
public static int getInvCount(int arr[], int n){
convert(arr, n);
for (int i = n - 1; i >= 0; i--) {
updateBIT(1, arr[i], 1, n);
for (int l = 1; l < 3; l++) {
updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n);
}
}
return getSum(3, n);
}
public static void main (String[] args){
int arr[] = {8, 4, 2, 1};
int n = arr.length;
System.out.print("Inversion Count After The Operation : "+getInvCount(arr, n));
}
}


### Output

Inversion Count After The Operation : 4


## Conclusion 