Count Inversions of size three in a given array

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.

C++

// A Simple C++ O(n^3)  program to count inversions of size 3
#include<bits/stdc++.h>
using namespace std;

// Returns counts of inversions of size three
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
int main()
{
    int arr[] = {8, 4, 2, 1};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "Inversion Count : " << getInvCount(arr, n);
    return 0;
}

Java

// A simple Java implementation  to count inversion 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-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

Python3

# A simple python O(n^3) program
# to count inversions of size 3

# Returns counts of inversions
# of size threee
def getInvCount(arr):
    n = len(arr)
    invcount = 0  #Initialize result
    for i in range(0,n-1):
        for j in range(i+1 , n):
                if arr[i] > arr[j]:
                    for k in range(j+1 , n):
                        if arr[j] > arr[k]:
                            invcount += 1
    return invcount

# Driver program to test above function
arr = [8 , 4, 2 , 1]
print ("Inversion Count : %d" %(getInvCount(arr)))

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// A simple C# implementation to
// count inversion of size 3
using System;
class GFG {

// returns count of inversion of size 3
static int getInvCount(int []arr, int n)
    {

        // initialize result
        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;
    }

    // Driver Code
    public static void Main()
    {
        int []arr = new int[] {8, 4, 2, 1};
        int n = arr.Length;
        Console.WriteLine("Inversion count : " +
                           getInvCount(arr, n));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// A O(n^2) PHP program to
// count inversions of size 3

// Returns count of
// inversions of size 3
function getInvCount($arr, $n)
{

    // Initialize result
    $invcount = 0;

    for ($i = 1; $i < $n - 1; $i++)
    {

        // Count all smaller elements
        // on right of arr[i]
        $small = 0;
        for($j = $i + 1; $j < $n; $j++)
            if ($arr[$i] > $arr[$j])
                $small++;

        // Count all greater elements
        // on left of arr[i]
        $great = 0;
        for($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 Code
    $arr = array(8, 4, 2, 1);
    $n = sizeof($arr);
    echo "Inversion Count : "
        , getInvCount($arr, $n);

// This code is contributed m_kit
?>

Javascript

<script>
// A simple Javascript implementation  to count inversion of size 3

    // returns count of inversion of size 3
    function getInvCount(arr, n)
    {
        let invcount = 0; // initialize result

        for(let i = 0 ; i < n - 2; i++)
        {
            for(let j = i + 1; j < n - 1; j++)
            {
                if(arr[i] > arr[j])
                {
                    for(let k = j + 1; k < n; k++)
                    {
                        if(arr[j] > arr[k])
                            invcount++;
                    }
                }
            }
        }
        return invcount;
    }

    // driver program to test above function
    let arr = [8, 4, 2, 1];
    let n = arr.length;
    document.write("Inversion count : " +
                    getInvCount(arr, n));

    // This code is contributed by rag2127


</script>

Output

Inversion Count : 4

Time complexity: O(n^3)
Auxiliary Space: O(1).

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.

C++

// A O(n^2) C++  program to count inversions of size 3
#include<bits/stdc++.h>
using namespace std;

// Returns count of inversions of size 3
int getInvCount(int arr[], int n)
{
    int invcount = 0;  // Initialize result

    for (int i=1; 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
int main()
{
    int arr[] = {8, 4, 2, 1};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "Inversion Count : " << getInvCount(arr, n);
    return 0;
}

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

Python3

# A O(n^2) Python3 program to
#  count inversions of size 3

# Returns count of inversions
# of size 3
def getInvCount(arr, n):

    # Initialize result
    invcount = 0

    for i in range(1,n-1):

        # Count all smaller elements
        # on right of arr[i]
        small = 0
        for j in range(i+1 ,n):
            if (arr[i] > arr[j]):
                small+=1

        # Count all greater elements
        # on left of arr[i]
        great = 0;
        for j in range(i-1,-1,-1):
            if (arr[i] < arr[j]):
                great+=1

        # 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
arr = [8, 4, 2, 1]
n = len(arr)
print("Inversion Count :",getInvCount(arr, n))

# This code is Contributed by Smitha Dinesh Semwal

C#

// A O(n^2) Java program to count inversions
// of size 3
using System;

public class Inversion {

    // returns count of inversion of size 3
    static 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()
    {

        int []arr = new int[] {8, 4, 2, 1};
        int n = arr.Length;
        Console.WriteLine("Inversion count : "
                       + getInvCount(arr, n));
    }
}

// This code has been contributed by anuj_67.

PHP

<?php
// A O(n^2) PHP program to count
// inversions of size 3

// Returns count of
// inversions of size 3
function getInvCount($arr, $n)
{
    // Initialize result
    $invcount = 0;

    for ($i = 1; $i < $n - 1; $i++)
    {
        // Count all smaller elements
        // on right of arr[i]
        $small = 0;
        for ($j = $i + 1; $j < $n; $j++)
            if ($arr[$i] > $arr[$j])
                $small++;

        // Count all greater elements
        // on left of arr[i]
        $great = 0;
        for ($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 Code
$arr = array (8, 4, 2, 1);
$n = sizeof($arr);
echo "Inversion Count : " ,
      getInvCount($arr, $n);

// This code is contributed by m_kit
?>

Javascript

<script>
// A O(n^2) Javascript  program to count inversions of size 3

    // returns count of inversion of size 3
    function getInvCount(arr, n)
    {
        let invcount = 0; // initialize result

        for (let i = 0 ; i < n - 1; i++)
        {
            // count all smaller elements on right of arr[i]
            let small = 0;
            for (let j = i + 1; j < n; j++)
                if (arr[i] > arr[j])
                    small++;

            // count all greater elements on left of arr[i]
            let great = 0;
            for (let 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
    let arr=[8, 4, 2, 1];
    let n = arr.length;
    document.write("Inversion count : " +getInvCount(arr, n));

    // This code is contributed by avanitrachhadiya2155
</script>

Output

Inversion Count : 4

Time Complexity: O(n^2)
Auxiliary Space: O(1).

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 :

1. 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.
2. 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].

Below is the code for the above approach.

C++

// C++ program to count inversions of size 3 using
// Binary Indexed Tree
#include <bits/stdc++.h>
using namespace std;

// It is beneficial to declare the 2D BIT globally
// since passing it into functions will create
// additional overhead
const int N = 100005;
int BIT[3][N] = { 0 };

// update function. "t" denotes the t'th Binary
// indexed tree
void updateBIT(int t, int i, int val, int n)
{
    // Traversing the t'th BIT
    while (i <= n) {
        BIT[t][i] = BIT[t][i] + val;
        i = i + (i & (-i));
    }
}

// function to get the sum.
// "t" denotes the t'th Binary indexed tree
int getSum(int t, int i)
{
    int res = 0;

    // Traversing the t'th BIT
    while (i > 0) {
        res = res + BIT[t][i];
        i = i - (i & (-i));
    }
    return res;
}

// Converts an array to an array with values from 1 to n
// and relative order of smaller and greater elements
// remains same.
void convert(int arr[], int n)
{
    // Create a copy of arr[] in temp and sort
    // the temp array in increasing order
    int temp[n];
    for (int i = 0; i < n; i++)
        temp[i] = arr[i];
    sort(temp, temp + n);

    // Traverse all array elements
    for (int i = 0; i < n; i++) {

        // lower_bound() Returns pointer to the
        // first element greater than or equal
        // to arr[i]
        arr[i] = lower_bound(temp, temp + n,
                        arr[i]) - temp + 1;
    }
}

// Returns count of inversions of size three
int getInvCount(int arr[], int n)
{
    // Convert arr[] to an array with values from
    // 1 to n and relative order of smaller and
    // greater elements remains same.
    convert(arr, n);

    // iterating over the converted array in
    // reverse order.
    for (int i = n - 1; i >= 0; i--) {

        // update the BIT for l = 1
        updateBIT(1, arr[i], 1, n);

        // update BIT for all other BITs
        for (int l = 1; l < 3; l++) {
            updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n);
        }
    }

    // final result
    return getSum(3, n);
}

// Driver program to test above function
int main()
{
    int arr[] = {8, 4, 2, 1};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "Inversion Count : " << getInvCount(arr, n);
    return 0;
}

// This code is contributed by Pushpesh Raj.

Java

// Java program to count inversions of size 3 using
// Binary Indexed Tree

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

class GFG {

// It is beneficial to declare the 2D BIT globally
// since passing it into functions will create
// additional overhead
static int N = 100005;
static int BIT[][] = new int[4][N];

// update function. "t" denotes the t'th Binary
// indexed tree
static void updateBIT(int t, int i, int val, int n)
{
    // Traversing the t'th BIT
    while (i <= n) {
        BIT[t][i] = BIT[t][i] + val;
        i = i + (i & (-i));
    }
}

// function to get the sum.
// "t" denotes the t'th Binary indexed tree
static int getSum(int t, int i)
{
    int res = 0;

    // Traversing the t'th BIT
    while (i > 0) {
        res = res + BIT[t][i];
        i = i - (i & (-i));
    }
    return res;
}

// Converts an array to an array with values from 1 to n
// and relative order of smaller and greater elements
// remains same.
static void convert(int arr[], int n)
{
    // Create a copy of arr[] in temp and sort
    // the temp array in increasing order
    int temp[]=new int[n];
    for (int i = 0; i < n; i++)
        temp[i] = arr[i];
    Arrays.sort(temp);

    // Traverse all array elements
    for (int i = 0; i < n; i++) {

        // lower_bound() Returns pointer to the
        // first element greater than or equal
        // to arr[i]
        arr[i] = Arrays.binarySearch(temp,arr[i]) + 1;
    }
}

// Returns count of inversions of size three
static int getInvCount(int arr[], int n)
{
    // Convert arr[] to an array with values from
    // 1 to n and relative order of smaller and
    // greater elements remains same.
    convert(arr, n);

    // iterating over the converted array in
    // reverse order.
    for (int i = n - 1; i >= 0; i--) {

        // update the BIT for l = 1
        updateBIT(1, arr[i], 1, n);

        // update BIT for all other BITs
        for (int l = 1; l < 3; l++) {
            updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n);
        }
    }

    // final result
    return getSum(3, n);
}

// Driver program to test above function
public static void main (String[] args)
{
    int arr[] = {8, 4, 2, 1};
    int n = arr.length;
    System.out.print("Inversion Count : "+getInvCount(arr, n));
}
}

// This code is contributed by Utkarsh

Python3

# Python program to count inversions of size 3 using
# Binary Indexed Tree

# It is beneficial to declare the 2D BIT globally
# since passing it into functions will create
# additional overhead
N = 100005
BIT = [[0 for x in range(N)] for y in range(3)]
for i in range(0,50):
  abc = [0]*50
  BIT.append(abc)

# update function. "t" denotes the t'th Binary
# indexed tree
def updateBIT(t, i, val, n):
  # Traversing the t'th BIT
  while i <= n:
    BIT[t][i] = BIT[t][i] + val
    i = i + (i & (-i))

# function to get the sum.
# "t" denotes the t'th Binary indexed tree
def getSum(t, i):
    res = 0

    # Traversing the t'th BIT
    while i > 0:
        res = res + BIT[t][i]
        i = i - (i & (-i))
    return res

# Converts an array to an array with values from 1 to n
# and relative order of smaller and greater elements
# remains same.
def convert(arr, n):
    # Create a copy of arr[] in temp and sort
    # the temp array in increasing order
    temp = [0 for x in range(n)]
    for i in range(n):
        temp[i] = arr[i]
    temp.sort()

    # Traverse all array elements
    for i in range(n):
        # lower_bound() Returns pointer to the
        # first element greater than or equal
        # to arr[i]
        arr[i] = temp.index(arr[i]) + 1

# Returns count of inversions of size three
def getInvCount(arr, n):
    # Convert arr[] to an array with values from
    # 1 to n and relative order of smaller and
    # greater elements remains same.
    convert(arr, n)

    # iterating over the converted array in
    # reverse order.
    for i in range(n - 1, -1, -1):
        # update the BIT for l = 1
        updateBIT(1, arr[i], 1, n)

        # update BIT for all other BITs
        for l in range(1, 3):
            updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n)

    # final result
    return getSum(3, n)

# Driver program to test above function
arr = [8, 4, 2, 1]
n = len(arr)
print("Inversion Count : ", getInvCount(arr, n))

# This code is contributed by ashishak__

Javascript

// JavaScript program to count inversions of size 3 using
// Binary Indexed Tree
const N = 100005;
let BIT = [[0], [0], [0]];

for (let i = 0; i < 10; i++) {
  let abc = [0];
  BIT.push(abc);
}

// update function. "t" denotes the t'th Binary
// indexed tree
function updateBIT(t, i, val, n) {
  // Traversing the t'th BIT
  while (i <= n) {
    if (!BIT[t][i]) BIT[t][i] = 0;
    BIT[t][i] = BIT[t][i] + val;
    i = i + (i & (-i));
  }
}

// function to get the sum.
// "t" denotes the t'th Binary indexed tree
function getSum(t, i) {
  let res = 0;

  // Traversing the t'th BIT
  while (i > 0) {
    res = res + BIT[t][i];
    i = i - (i & (-i));
  }
  return res;
}

// Converts an array to an array with values from 1 to n
// and relative order of smaller and greater elements
// remains same.
function convert(arr, n) {
  // Create a copy of arr[] in temp and sort
  // the temp array in increasing order
  let temp = [];
  for (let i = 0; i < n; i++)
    temp[i] = arr[i];
  temp.sort();

  // Traverse all array elements
  for (let i = 0; i < n; i++) {

    // lower_bound() Returns pointer to the
    // first element greater than or equal
    // to arr[i]
    arr[i] = temp.indexOf(arr[i]) + 1;
  }
}

// Returns count of inversions of size three
function getInvCount(arr, n) {
  // Convert arr[] to an array with values from
  // 1 to n and relative order of smaller and
  // greater elements remains same.
  convert(arr, n);

  // iterating over the converted array in
  // reverse order.
  for (let i = n - 1; i >= 0; i--) {

    // update the BIT for l = 1
    updateBIT(1, arr[i], 1, n);

    // update BIT for all other BITs
    for (let l = 1; l < 3; l++) {
      updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n);
    }
  }

  // final result
  return getSum(3, n);
}

// Driver program to test above function

let arr = [8, 4, 2, 1];
let n = arr.length;
console.log("Inversion Count : " + getInvCount(arr, n));

// This code is contributed by akashish__.

C#

using System;

class GFG
{
    // It is beneficial to declare the 2D BIT globally
    // since passing it into functions will create
    // additional overhead
    static int N = 100005;
    static int[,] BIT = new int[4, N];

    // update function. "t" denotes the t'th Binary
    // indexed tree
    static void updateBIT(int t, int i, int val, int n)
    {
        // Traversing the t'th BIT
        while (i <= n)
        {
            BIT[t, i] = BIT[t, i] + val;
            i = i + (i & (-i));
        }
    }

    // function to get the sum.
    // "t" denotes the t'th Binary indexed tree
    static int getSum(int t, int i)
    {
        int res = 0;

        // Traversing the t'th BIT
        while (i > 0)
        {
            res = res + BIT[t, i];
            i = i - (i & (-i));
        }
        return res;
    }

    // Converts an array to an array with values from 1 to n
    // and relative order of smaller and greater elements
    // remains same.
    static void convert(int[] arr, int n)
    {
        // Create a copy of arr[] in temp and sort
        // the temp array in increasing order
        int[] temp = new int[n];
        for (int i = 0; i < n; i++)
            temp[i] = arr[i];
        Array.Sort(temp);

        // Traverse all array elements
        for (int i = 0; i < n; i++)
        {
            // lower_bound() Returns pointer to the
            // first element greater than or equal
            // to arr[i]
            arr[i] = Array.BinarySearch(temp, arr[i]) + 1;
        }
    }

    // Returns count of inversions of size three
    static int getInvCount(int[] arr, int n)
    {
        // Convert arr[] to an array with values from
        // 1 to n and relative order of smaller and
        // greater elements remains same.
        convert(arr, n);

        // iterating over the converted array in
        // reverse order.
        for (int i = n - 1; i >= 0; i--)
        {
            // update the BIT for l = 1
            updateBIT(1, arr[i], 1, n);

            // update BIT for all other BITs
            for (int l = 1; l < 3; l++)
            {
                updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n);
            }
        }

        // final result
        return getSum(3, n);
    }

    // Driver program to test above function
    static void Main(string[] args)
    {
        int[] arr = { 8, 4, 2, 1 };
        int n = arr.Length;
        Console.Write("Inversion Count : " + getInvCount(arr, n));
    }
}

Output

Inversion Count : 4

Time Complexity: O(n*log(n))
Auxiliary Space: O(n).