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Count inversions in an array | Set 4 ( Using Trie )
  • Difficulty Level : Expert
  • Last Updated : 02 Dec, 2020
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Inversion Count for an array indicates – how far (or close) the array is from being sorted. If the array is already sorted then inversion count is 0. If the array is sorted in reverse order that inversion count is the maximum.
 

Two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j. 
For simplicity, we may assume that all elements are unique. 
 

So, our task is to count the number of inversions in the array. That is the number of pair of elements (a[i], a[j]) such that: 
 

  • a[i] > a[j] and,
  • i < j.

Example
 

Input: arr[] = {8, 4, 2, 1}
Output: 6
Given array has six inversions (8, 4), (4, 2),
(8, 2), (8, 1), (4, 1), (2, 1). 

Input: arr[] = { 1, 20, 6, 4, 5 }    
Output: 5

 



 

  1. Naive and Modified Merge Sort
  2. Using AVL Tree
  3. Using BIT

Approach: 
We will iterate backwards in the array and store each element into the Trie. To store a number in Trie we 
have to break the number into its binary form and If the bit is 0 then it signifies we store that bit into the left pointer of the current node and if it is 1 we will store it into the right pointer of the current node and correspondingly change the current node. We will also maintain the count which signifies how many numbers follow the same path till that node.
Structure of Node of the Trie 
 

struct node{
  int count; 
  node* left;
  node* right;
};

At any point, while we are storing the bits, we happen to move to the right pointer (i.e the bit is 1) we will check if the left child exists then this means there are numbers which are smaller than the current number who are already been stored into the Trie, these numbers are only the inversion count so we will add these to the count. 
Below is the implementation of the approach 
 

C++




// C++ implementation
#include <iostream>
using namespace std;
 
// Structure of the node
struct Node {
    int count;
    Node* left;
    Node* right;
};
 
// function to initialize
// new node
Node* makeNewNode()
{
    Node* temp = new Node;
    temp->count = 1;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}
 
// Insert element in trie
void insertElement(int num,
                   Node* root,
                   int& ans)
{
    // Converting the number
    // into binary form
    for (int i = 63; i >= 0; i--) {
        // Checking if the i-th
        // bit ios set or not
        int a = (num & (1 << i));
 
        // If the bit is 1
        if (a != 0) {
            // if the bit is 1 that means
            // we have to go to the right
            // but we also checks if left
            // pointer  exists i.e there is
            // at least a number smaller than
            // the current number already in
            // the trie we add that count
            // to ans
            if (root->left != NULL)
                ans += root->left->count;
 
            // If right pointer is not NULL
            // we just iterate to that
            // position and increment the count
            if (root->right != NULL) {
                root = root->right;
                root->count += 1;
            }
 
            // If right is NULL we add a new
            // node over there and initialize
            // the count with 1
            else {
                Node* temp = makeNewNode();
                root->right = temp;
                root = root->right;
            }
        }
 
        // if the bit is 0
        else {
            // We have to iterate to left,
            // we first check if left
            // exists? if yes then change
            // the root and the count
            if (root->left != NULL) {
                root = root->left;
                root->count++;
            }
 
            // otherwise we create
            // the left node
            else {
                Node* temp = makeNewNode();
                root->left = temp;
                root = root->left;
            }
        }
    }
}
 
// function to count
// the inversions
int getInvCount(int arr[], int n)
{
    Node* head = makeNewNode();
    int ans = 0;
    for (int i = n - 1; i >= 0; i--) {
        // inserting each element in Trie
        insertElement(arr[i],
                      head,
                      ans);
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 8, 4, 2, 1 };
    int n = sizeof(arr) / sizeof(int);
 
    cout << "Number of inversions are : "
         << getInvCount(arr, n);
    return 0;
}

Java




// Java implementation of above idea
import java.util.*;
 
class GFG
{
 
// Structure of the node
static class Node
{
    int count;
    Node left;
    Node right;
};
static int ans;
 
// function to initialize
// new node
static Node makeNewNode()
{
    Node temp = new Node();
    temp.count = 1;
    temp.left = null;
    temp.right = null;
    return temp;
}
 
// Insert element in trie
static void insertElement(int num,
                Node root)
{
    // Converting the number
    // into binary form
    for (int i = 63; i >= 0; i--)
    {
        // Checking if the i-th
        // bit ios set or not
        int a = (num & (1 << i));
 
        // If the bit is 1
        if (a != 0)
        {
            // if the bit is 1 that means
            // we have to go to the right
            // but we also checks if left
            // pointer exists i.e there is
            // at least a number smaller than
            // the current number already in
            // the trie we add that count
            // to ans
            if (root.left != null)
                ans += root.left.count;
 
            // If right pointer is not null
            // we just iterate to that
            // position and increment the count
            if (root.right != null)
            {
                root = root.right;
                root.count += 1;
            }
 
            // If right is null we add a new
            // node over there and initialize
            // the count with 1
            else
            {
                Node temp = makeNewNode();
                root.right = temp;
                root = root.right;
            }
        }
 
        // if the bit is 0
        else
        {
            // We have to iterate to left,
            // we first check if left
            // exists? if yes then change
            // the root and the count
            if (root.left != null)
            {
                root = root.left;
                root.count++;
            }
 
            // otherwise we create
            // the left node
            else
            {
                Node temp = makeNewNode();
                root.left = temp;
                root = root.left;
            }
        }
    }
}
 
// function to count
// the inversions
static int getInvCount(int arr[], int n)
{
    Node head = makeNewNode();
    ans = 0;
    for (int i = n - 1; i >= 0; i--)
    {
        // inserting each element in Trie
        insertElement(arr[i],
                    head);
    }
 
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 8, 4, 2, 1 };
    int n = arr.length;
 
    System.out.print("Number of inversions are : "
        + getInvCount(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation
  
# Structure of the node
class Node:
     
    def __init__(self):
         
        self.left = self.right = None
        self.count = 1
 
# function to initialize
# new node
def makeNewNode():
     
    temp = Node()
    return temp
 
# Insert element in trie
def insertElement(num, root, ans):
 
    # Converting the number
    # into binary form
    for i in range(63, -1, -1):
     
        # Checking if the i-th
        # bit ios set or not
        a = (num & (1 << i));
  
        # If the bit is 1
        if (a != 0):
       
            # if the bit is 1 that means
            # we have to go to the right
            # but we also checks if left
            # pointer  exists i.e there is
            # at least a number smaller than
            # the current number already in
            # the trie we add that count
            # to ans
            if (root.left != None):
                ans += root.left.count;
  
            # If right pointer is not None
            # we just iterate to that
            # position and increment the count
            if (root.right != None):
                root = root.right;
                root.count += 1;
             
            # If right is None we add a new
            # node over there and initialize
            # the count with 1
            else:
                temp = makeNewNode();
                root.right = temp;
                root = root.right;
      
        # if the bit is 0
        else:
           
            # We have to iterate to left,
            # we first check if left
            # exists? if yes then change
            # the root and the count
            if (root.left != None):
                root = root.left;
                root.count += 1
  
            # otherwise we create
            # the left node
            else:
                temp = makeNewNode();
                root.left = temp;
                root = root.left;
    return ans
 
# function to count
# the inversions
def getInvCount(arr, n):
 
    head = makeNewNode();
    ans = 0;
     
    for i in range(n - 1 ,-1, -1):
     
        # inserting each element in Trie
        ans = insertElement(arr[i], head, ans);
  
    return ans;
 
# Driver Code
if __name__=='__main__':
 
    arr = [ 8, 4, 2, 1 ]
     
    n = len(arr)
     
    print("Number of inversions are : " + str(getInvCount(arr, n)))
 
    # This code is contributed by rutvik_56

C#




// C# implementation of above idea
using System;
 
class GFG
{
 
// Structure of the node
public class Node
{
    public int count;
    public Node left;
    public Node right;
};
static int ans;
 
// function to initialize
// new node
static Node makeNewNode()
{
    Node temp = new Node();
    temp.count = 1;
    temp.left = null;
    temp.right = null;
    return temp;
}
 
// Insert element in trie
static void insertElement(int num,
                         Node root)
{
    // Converting the number
    // into binary form
    for (int i = 63; i >= 0; i--)
    {
        // Checking if the i-th
        // bit ios set or not
        int a = (num & (1 << i));
 
        // If the bit is 1
        if (a != 0)
        {
            // if the bit is 1 that means
            // we have to go to the right
            // but we also checks if left
            // pointer exists i.e there is
            // at least a number smaller than
            // the current number already in
            // the trie we add that count
            // to ans
            if (root.left != null)
                ans += root.left.count;
 
            // If right pointer is not null
            // we just iterate to that
            // position and increment the count
            if (root.right != null)
            {
                root = root.right;
                root.count += 1;
            }
 
            // If right is null we add a new
            // node over there and initialize
            // the count with 1
            else
            {
                Node temp = makeNewNode();
                root.right = temp;
                root = root.right;
            }
        }
 
        // if the bit is 0
        else
        {
            // We have to iterate to left,
            // we first check if left
            // exists? if yes then change
            // the root and the count
            if (root.left != null)
            {
                root = root.left;
                root.count++;
            }
 
            // otherwise we create
            // the left node
            else
            {
                Node temp = makeNewNode();
                root.left = temp;
                root = root.left;
            }
        }
    }
}
 
// function to count the inversions
static int getInvCount(int []arr, int n)
{
    Node head = makeNewNode();
    ans = 0;
    for (int i = n - 1; i >= 0; i--)
    {
        // inserting each element in Trie
        insertElement(arr[i], head);
    }
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 8, 4, 2, 1 };
    int n = arr.Length;
 
    Console.Write("Number of inversions are : " +
                            getInvCount(arr, n));
}
}
 
// This code is contributed by 29AjayKumar
Output: 
Number of inversions are : 6

 

Time Complexity: O(Nlog(N))
Auxiliary SpaceO(Nlog(N))
 

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