Pre-requisite: Policy based data structure Given an array arr[], the task is to find the number of inversions for each element of the array.
Inversion Count: for an array indicates – how far (or close) the array is from being sorted. If the array is already sorted then the inversion count is 0. If the array is sorted in the reverse order then the inversion count is the maximum. Formally, Number of indices
and
such that
and
.
and 
such that ![arr[i] > arr[j]](https://neveropen.co.uk/wp-content/uploads/2023/11/wardslaus_rubhabha_quicklatex.com-add979df51b9e2787bce263d492d4bae_l3.png "Rendered by QuickLaTeX.com")
and 
.Examples:
Input: {5, 2, 3, 2, 3, 8, 1} Output: {0, 1, 1, 2, 1, 0, 6} Explanation: Inversion count for each elements – Element at index 0: There are no elements with less index than 0, which is greater than arr[0]. Element at index 1: There is one element with less index than 1, which is greater than 2. That is 5. Element at index 2: There is one element with less index than 2, which is greater than 3. That is 5. Element at index 3: There are two elements with less index than 3, which is greater than 2. That is 5, 3. Element at index 4: There is one element with less index than 4, which is greater than 3. That is 5. Element at index 5: There are no elements with less index than 5, which is greater than 8. Element at index 6: There are six elements with less index than 6, which is greater than 1. That is 5, 2, 3, 2, 3 Input: arr[] = {3, 2, 1} Output: {0, 1, 2}
Approach:
- Create a policy based data structure of type pair.
- Iterate the given array and perform the following steps –
- Apply order_of_key({X, N+1}) for each element X where N is the size of array. Note: order_of_key is nothing but lower_bound. Also, we used N+1 because it is greater than all the indices in the array.
- Let order_of_key comes out to be l, then the inversion count for current element will be equal to
which is ultimately the count of elements smaller than X and came before X in the array. - Insert the current element X along with its index in the policy-based data structure St. The index is inserted along with each element for its unique identification in the set and to deal with duplicates.
which is ultimately the count of elements smaller than X and came before X in the array.Below is the implementation of the above approach:
C++
// C++ implementation to find the// Inversion Count using Policy// Based Data Structure#include <bits/stdc++.h>// Header files for policy based// data structure which are// to be included#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>using namespace __gnu_pbds;using namespace std;typedef tree<pair<int, int>, null_type, less<pair<int, int> >, rb_tree_tag, tree_order_statistics_node_update> new_data_set;// Function to find the inversion count// of the elements of the arrayvoid inversionCount(int arr[], int n){ int ans[n]; // Making a new policy based data // structure which will // store pair<int, int> new_data_set St; // Loop to iterate over the elements // of the array for (int i = 0; i < n; i++) { // Now to find lower_bound of // the element X, we will use pair // as {x, n+1} to cover all the // elements and even the duplicates int cur = St.order_of_key({ arr[i], n + 1 }); ans[i] = St.size() - cur; // Store element along with index St.insert({ arr[i], i }); } for (int i = 0; i < n; i++) { cout << ans[i] << " "; } cout << "\n";}// Driver Codeint main(){ int arr[] = { 5, 2, 3, 2, 3, 8, 1 }; int n = sizeof(arr) / sizeof(int); // Function Call inversionCount(arr, n); return 0;} |
Java
import java.util.*;import java.io.*;import java.lang.*;import java.util.AbstractMap.SimpleEntry;import java.util.Map.Entry;// Header files for policy based data structure which are to be includedimport java.util.TreeSet;public class GFG{ public static void inversionCount(int[] arr, int n) { int[] ans = new int[n]; // Making a new policy based data structure which will store pair<int, int> TreeSet<Entry<Integer, Integer>> St = new TreeSet<>(new Comparator<Entry<Integer, Integer>>() { @Override public int compare(Entry<Integer, Integer> a, Entry<Integer, Integer> b) { return a.getKey().compareTo(b.getKey()); } }); // Loop to iterate over the elements of the array for (int i = 0; i < n; i++) { // Now to find lower_bound of the element X, we will use pair as {x, n+1} to cover all the // elements and even the duplicates int cur = St.headSet(new SimpleEntry<Integer, Integer>(arr[i], n+1)).size(); ans[i] = St.size() - cur; // Store element along with index St.add(new SimpleEntry<Integer, Integer>(arr[i], i)); } for (int i = 0; i < n; i++) { System.out.print(ans[i] + " "); } System.out.print("\n"); } public static void main(String[] args) { int[] arr = { 5, 2, 3, 2, 3, 8, 1 }; int n = arr.length; // Function Call inversionCount(arr, n); }} |
Python3
# Python implementation to find the# Inversion Count using Policy# Based Data Structure# Import required librariesfrom collections import defaultdictfrom bisect import bisect_right# Function to find the inversion count# of the elements of the arraydef inversionCount(arr, n): ans = [0] * n # Making a new dictionary to store index # of every element index_dict = defaultdict(list) # Loop to iterate over the elements # of the array for i in range(n): # Find the index of the current element in # the sorted list of elements so far cur = bisect_right(index_dict[arr[i]], i) ans[i] = i - cur # Store the current element's index index_dict[arr[i]].append(i) for i in range(n): print(ans[i], end=' ') print()# Driver Codeif __name__ == '__main__': arr = [5, 2, 3, 2, 3, 8, 1] n = len(arr) # Function Call inversionCount(arr, n) |
Javascript
// Function to find the inversion count// of the elements of the arrayfunction inversionCount(arr, n) { let ans = new Array(n).fill(0); // Making a new object to store index // of every element let index_dict = {}; // Loop to iterate over the elements // of the array for (let i = 0; i < n; i++) { // Find the index of the current element in // the sorted list of elements so far if (arr[i] in index_dict) { let cur = index_dict[arr[i]].filter((index) => index < i).length; ans[i] = i - cur; } else { ans[i] = i; } // Store the current element's index if (arr[i] in index_dict) { index_dict[arr[i]].push(i); } else { index_dict[arr[i]] = [i]; } } console.log(ans.join(' '));}// Driver Codelet arr = [5, 2, 3, 2, 3, 8, 1];let n = arr.length;// Function CallinversionCount(arr, n);// This code is contributed by Aditya Sharma |
0 1 1 2 1 0 6
Time Complexity: 
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