🔥 ข่าวด่วน: web//https://www.geeksforgeeks.org/count inversions in an array set using trie

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive and Modified Merge Sort
Using AVL Tree
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 

Output:

Number of inversions are : 6

Time Complexity: $O(Nlog(N))$
Auxiliary Space: $O(Nlog(N))$

My Personal Notes

Nov	DEC	Jan
	25
2018	2019	2020

Count inversions in an array | Set 4 ( Using Trie )

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Recommended Posts: