Counting inversions in an array using segment tree

Last Updated : 12 Sep, 2022

Given an array of integers arr, the task is to count the number of inversions in the array.
If A[i] > A[j] and i < j then the pair (A[i], A[j]) is part of an inversion.

Examples:

Input: arr[] = {8, 4, 2, 1}
Output: 6

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

Approach:

Build a segment tree where each node will represent the total numbers present in the range of that node.
Let's say the range of any node is [i, j], then the node will contain the count of numbers which are greater than or equal to i and less than or equal to j.
Leaf nodes will only be either 1 or 0 since range of the node will be 1.
Iterate through the array, let the number present at the index i is a[i]. We will find how many numbers are present in the segment tree in the range [a[i]+1, max] where max is the maximum element of the array and add it to the answer variable.
Then we will insert that number in the segment tree and continue till the last index of the array. This way for each element we are adding the numbers which appear before that element and are greater than that element i.e. they form an inversion pair.

Below is the implementation of the above approach:

C++

// C++ program to count the inversions
// present in the array using segment tree
#include "algorithm"
#include "cstring"
#include "iostream"
using namespace std;

// Function to update segment tree
// i.e. to insert the element
void update(int* Tree, int index, int s, int e, int num)
{
    // Leaf node condition
    if (s == num and e == num) {
        Tree[index] = 1;
        return;
    }

    // No overlap condition
    if (num < s or num > e) {
        return;
    }

    // Else call on both the children nodes
    int mid = (s + e) >> 1;
    update(Tree, 2 * index, s, mid, num);
    update(Tree, 2 * index + 1, mid + 1, e, num);

    Tree[index] = Tree[2 * index] + Tree[2 * index + 1];
}

// Function to count the total numbers
// present in the range [a[i]+1, mx]
int query(int* Tree, int index, int s,
          int e, int qs, int qe)
{
    // Complete overlap
    if (qs <= s and e <= qe) {
        return Tree[index];
    }

    // No Overlap
    if (s > qe or e < qs) {
        return 0;
    }

    int mid = (s + e) >> 1;

    return query(Tree, 2 * index, s,
                 mid, qs, qe)
           + query(Tree, 2 * index + 1,
                   mid + 1, e, qs, qe);
}

// Driver code
int main(int argc, char const* argv[])
{
    int arr[] = { 8, 4, 2, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Maximum element present in the array.
    int mx = *max_element(arr, arr + n);

    // Segment Tree
    int Tree[6 * mx];

    // Initialize every node
    // of segment tree to 0
    memset(Tree, 0, sizeof(Tree));

    int answer = 0;

    for (int i = 0; i < n; ++i) {

        // add the count of inversion pair
        // formed with the element a[i] and the
        // elements appearing before the index i.
        answer += query(Tree, 1, 1, mx, arr[i] + 1, mx);

        // Insert the a[i] in the segment tree
        update(Tree, 1, 1, mx, arr[i]);
    }

    cout << answer;
    return 0;
}

Java

// Java program to count the inversions 
// present in the array using segment tree 
import java.io.*;
import java.util.Arrays;
class GFG 
{

  // Function to update segment tree 
  // i.e. to insert the element 
  static void update(int[] Tree, int index, 
                     int s, int e, int num)
  {

    // Leaf node condition 
    if (s == num && e == num)
    { 
      Tree[index] = 1; 
      return; 
    }

    // No overlap condition
    if (num < s || num > e)
    {
      return; 
    }

    // Else call on both the children nodes 
    int mid = (s + e) >> 1;
    update(Tree, 2 * index, s, mid, num); 
    update(Tree, 2 * index + 1, mid + 1, e, num); 

    Tree[index] = Tree[2 * index] + Tree[2 * index + 1]; 
  }

  // Function to count the total numbers 
  // present in the range [a[i]+1, mx] 
  static int query(int[] Tree, int index, int s,
                   int e, int qs, int qe) 
  {
    // Complete overlap 
    if (qs <= s && e <= qe)
    {
      return Tree[index];
    }

    // No Overlap 
    if (s > qe || e < qs)
    {
      return 0; 
    }
    int mid = (s + e) >> 1; 
    return query(Tree, 2 * index, s, mid, qs, qe) + 
      query(Tree, 2 * index + 1, mid + 1, e, qs, qe);
  }

  // Driver code
  public static void main (String[] args)
  {
    int arr[] = { 8, 4, 2, 1 }; 
    int n = arr.length;

    // Maximum element present in the array. 
    int mx = Arrays.stream(arr).max().getAsInt(); 

    // Segment Tree 
    int[] Tree = new int[6 * mx];
    int answer = 0; 
    for (int i = 0; i < n; ++i)
    {

      // add the count of inversion pair 
      // formed with the element a[i] and the 
      // elements appearing before the index i. 
      answer += query(Tree, 1, 1, mx, arr[i] + 1, mx); 

      // Insert the a[i] in the segment tree 
      update(Tree, 1, 1, mx, arr[i]); 
    }    
    System.out.println(answer);
  }
}

// This code is contributed by rag2127

Python3

# Python3 program to count the inversions
# present in the array using segment tree

# Function to update segment tree
# i.e. to insert the element
def update(Tree, index, s, e, num):
    
    # Leaf node condition
    if (s == num and e == num):
        Tree[index] = 1
        return

    # No overlap condition
    if (num < s or num > e):
        return

    # Else call on both the children nodes
    mid = (s + e) >> 1
    update(Tree, 2 * index, s, mid, num)
    update(Tree, 2 * index + 1, mid + 1, e, num)

    Tree[index] = Tree[2 * index] + Tree[2 * index + 1]

# Function to count the total numbers
# present in the range [a[i]+1, mx]
def query(Tree,index, s,e, qs, qe):
    
    # Complete overlap
    if (qs <= s and e <= qe):
        return Tree[index]

    # No Overlap
    if (s > qe or e < qs):
        return 0

    mid = (s + e) >> 1

    return query(Tree, 2 * index, s,mid, qs, qe) +\
           query(Tree, 2 * index + 1, mid + 1, e, qs, qe)

# Driver code
if __name__ == '__main__':
    arr = [8, 4, 2, 1]
    n = len(arr)

    # Maximum element present in the array.
    mx = max(arr)

    # Segment Tree
    Tree = [0] * (6 * mx)

    # Initialize every node
    # of segment tree to 0
    answer = 0

    for i in range(n):

        # add the count of inversion pair
        # formed with the element a[i] and the
        # elements appearing before the index i.
        answer += query(Tree, 1, 1, mx, arr[i] + 1, mx)

        # Insert the a[i] in the segment tree
        update(Tree, 1, 1, mx, arr[i])

    print(answer)

# This code is contributed by Mohit Kumar

using System;
using System.Linq;

class GFG
{

  // Function to update segment tree 
  // i.e. to insert the element 
  static void update(int[] Tree, int index, 
                     int s, int e, int num)
  {

    // Leaf node condition 
    if (s == num && e == num)
    { 
      Tree[index] = 1; 
      return; 
    }

    // No overlap condition
    if (num < s || num > e)
    {
      return; 
    }

    // Else call on both the children nodes 
    int mid = (s + e) >> 1;
    update(Tree, 2 * index, s, mid, num); 
    update(Tree, 2 * index + 1, mid + 1, e, num);  
    Tree[index] = Tree[2 * index] + Tree[2 * index + 1]; 
  }

  // Function to count the total numbers 
  // present in the range [a[i]+1, mx] 
  static int query(int[] Tree, int index,
                   int s, int e, int qs, int qe) 
  {

    // Complete overlap 
    if (qs <= s && e <= qe)
    {
      return Tree[index];
    }

    // No Overlap 
    if (s > qe || e < qs)
    {
      return 0; 
    }
    int mid = (s + e) >> 1; 
    return query(Tree, 2 * index, s, mid, qs, qe) +
      query(Tree, 2 * index + 1, mid + 1, e, qs, qe);
  }


  // Driver code
  static public void Main ()
  {
    int[] arr = { 8, 4, 2, 1 }; 
    int n = arr.Length;

    // Maximum element present in the array. 
    int mx =  arr.Max();

    // Segment Tree 
    int[] Tree = new int[6 * mx];
    int answer = 0; 
    for (int i = 0; i < n; ++i)
    {

      // add the count of inversion pair 
      // formed with the element a[i] and the 
      // elements appearing before the index i. 
      answer += query(Tree, 1, 1, mx, arr[i] + 1, mx); 

      // Insert the a[i] in the segment tree 
      update(Tree, 1, 1, mx, arr[i]); 
    }   
    Console.WriteLine(answer);
  }
}

// This code is contributed by avanitrachhadiya2155

JavaScript

<script>

// JavaScript program to count the inversions
// present in the array using segment tree

 // Function to update segment tree
  // i.e. to insert the element
function update(Tree,index,s,e,num)
{
    // Leaf node condition
    if (s == num && e == num)
    {
      Tree[index] = 1;
      return;
    }
 
    // No overlap condition
    if (num < s || num > e)
    {
      return;
    }
 
    // Else call on both the children nodes
    let mid = (s + e) >> 1;
    update(Tree, 2 * index, s, mid, num);
    update(Tree, 2 * index + 1, mid + 1, e, num);
 
    Tree[index] = Tree[2 * index] + Tree[2 * index + 1];
  }
 
  // Function to count the total numbers
  // present in the range [a[i]+1, mx]
  function query(Tree, index, s,
                    e,  qs,  qe)
  {
    // Complete overlap
    if (qs <= s && e <= qe)
    {
      return Tree[index];
    }
 
    // No Overlap
    if (s > qe || e < qs)
    {
      return 0;
    }
    let mid = (s + e) >> 1;
    return query(Tree, 2 * index, s, mid, qs, qe) +
      query(Tree, 2 * index + 1, mid + 1, e, qs, qe);
}

// Driver code

let arr=[ 8, 4, 2, 1 ];
let n = arr.length;

// Maximum element present in the array.
    let mx = Math.max(...arr);
 
    // Segment Tree
    let Tree = new Array(6 * mx);
    for(let i=0;i<Tree.length;i++)
    {
        Tree[i]=0;
    }
    let answer = 0;
    for (let i = 0; i < n; ++i)
    {
 
      // add the count of inversion pair
      // formed with the element a[i] and the
      // elements appearing before the index i.
      answer += query(Tree, 1, 1, mx, arr[i] + 1, mx);
 
      // Insert the a[i] in the segment tree
      update(Tree, 1, 1, mx, arr[i]);
    }   
    document.write(answer);
    

// This code is contributed by unknown2108

</script>

Output

Count inversions of size k in a given array

imdhruvgupta

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