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Counting Inversions using Ordered Set and GNU C++ PBDS

  • Difficulty Level : Medium
  • Last Updated : 26 Jul, 2021

Given an array arr[] of N integers. The task is to find the number of inversion. Two elements arr[i] and arr[j] form an inversion if arr[i] > arr[j] and i < j

Examples 

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Input: arr[] = {8, 4, 2, 1} 
Output:
Explanation: 
Given array has six inversions:  



  • (8, 4): arr[0] > arr[1] and 0 < 1
  • (8, 2): arr[0] > arr[2] and 0 < 2
  • (8, 1): arr[0] > arr[3] and 0 < 3
  • (4, 2): arr[1] > arr[2] and 1 < 2
  • (4, 1): arr[1] > arr[3] and 1 < 3
  • (2, 1): arr[2] > arr[3] and 2 < 3

Input: arr[] = {2, 3} 
Output:
Explanation: 
There is no such pair exists such that arr[i] > arr[j] and i < j.
 

We have already discussed below approaches:  

In this post, we will be discussing an approach using Ordered Set and GNU C++ PBDS.

Approach: 
We will be using the function order_of_key(K) which returns number of elements strictly smaller than K in log N time.  

  1. Insert the first element of the array in the Ordered_Set.
  2. For all the remaining element in arr[] do the following: 
    • Insert the current element in the Ordered_Set.
    • Find the number of element strictly less than current element + 1 in Ordered_Set using function order_of_key(arr[i]+1).
    • The difference between size of Ordered_Set and order_of_key(current_element + 1) will given the inversion count for the current element.

For Example:

arr[] = {8, 4, 2, 1}
Ordered_Set S = {8}
For remaining element in arr[]:
At index 1, the element is  4
S = {4, 8}
key = order_of_key(5) = 1
The difference between size of S and key gives the total 
number of inversion count for that current element.
inversion_count = S.size() - key =  2 - 1 = 1
Inversion Pairs are: (8, 4)

At index 2, the element is  2
S = {2, 4, 8}
key = order_of_key(3) = 1
inversion_count = S.size() - key =  3 - 1 = 2
Inversion Pairs are: (8, 2) and (4, 2)

At index 3, the element is 1
S = {1, 2, 4, 8}
key = order_of_key(2) = 1
inversion_count = S.size() - key =  4 - 1 = 3
Inversion Pairs are: (8, 1), (4, 1) and (2, 1)

Total inversion count = 1 + 2 + 3 = 6

Below is the implementation of the above approach: 

C++




// Ordered set in GNU C++ based
// approach for inversion count
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
 
// Ordered Set Tree
typedef tree<int, null_type, less_equal<int>,
             rb_tree_tag,
             tree_order_statistics_node_update>
    ordered_set;
 
// Returns inversion count in
// arr[0..n-1]
int getInvCount(int arr[], int n)
{
    int key;
    // Initialise the ordered_set
    ordered_set set1;
 
    // Insert the first
    // element in set
    set1.insert(arr[0]);
 
    // Initialise inversion
    // count to zero
    int invcount = 0;
 
    // Finding the inversion
    // count for current element
    for (int i = 1; i < n; i++) {
        set1.insert(arr[i]);
 
        // Number of elements strictly
        // less than arr[i]+1
        key = set1.order_of_key(arr[i] + 1);
 
        // Difference between set size
        // and key will give the
        // inversion count
        invcount += set1.size() - key;
    }
    return invcount;
}
 
// Driver's Code
int main()
{
    int arr[] = { 8, 4, 2, 1 };
    int n = sizeof(arr) / sizeof(int);
 
    // Function call to count
    // inversion
    cout << getInvCount(arr, n);
    return 0;
}
Output: 
6

 

Time Complexity: O(Nlog N)
 




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