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
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 nodestruct Node { int count; Node* left; Node* right;};// function to initialize// new nodeNode* makeNewNode(){ Node* temp = new Node; temp->count = 1; temp->left = NULL; temp->right = NULL; return temp;}// Insert element in trievoid 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 inversionsint 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 Codeint 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 ideaimport java.util.*;class GFG{// Structure of the nodestatic class Node { int count; Node left; Node right;};static int ans;// function to initialize// new nodestatic Node makeNewNode(){ Node temp = new Node(); temp.count = 1; temp.left = null; temp.right = null; return temp;}// Insert element in triestatic 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 inversionsstatic 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 Codepublic 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 nodeclass Node: def __init__(self): self.left = self.right = None self.count = 1# function to initialize# new nodedef makeNewNode(): temp = Node() return temp# Insert element in triedef 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 inversionsdef 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 Codeif __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 ideausing System;class GFG{// Structure of the nodepublic class Node { public int count; public Node left; public Node right;};static int ans;// function to initialize// new nodestatic Node makeNewNode(){ Node temp = new Node(); temp.count = 1; temp.left = null; temp.right = null; return temp;}// Insert element in triestatic 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 inversionsstatic 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 Codepublic 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 |
Javascript
<script> // JavaScript implementation of above idea// Structure of the nodeclass Node { constructor() { this.count = 0; this.left = null; this.right = null; }};var ans = 0;// function to initialize// new nodefunction makeNewNode(){ var temp = new Node(); temp.count = 1; temp.left = null; temp.right = null; return temp;}// Insert element in triefunction insertElement(num, root){ // Converting the number // into binary form for (var i = 63; i >= 0; i--) { // Checking if the i-th // bit ios set or not var 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 { var 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 { var temp = makeNewNode(); root.left = temp; root = root.left; } } }}// function to count the inversionsfunction getInvCount(arr, n){ var head = makeNewNode(); ans = 0; for (var i = n - 1; i >= 0; i--) { // inserting each element in Trie insertElement(arr[i], head); } return ans;}// Driver Codevar arr = [8, 4, 2, 1];var n = arr.length;document.write("Number of inversions are : " + getInvCount(arr, n));</script> |
Output:
Number of inversions are : 6
Time Complexity: 
Auxiliary Space:
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