# Counting Inversions using Ordered Set and GNU C++ PBDS

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

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

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

Input: arr[] = {2, 3}
Output: 0
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.
3. 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:

 `// Ordered set in GNU C++ based ` `// approach for inversion count ` `#include ` `#include ` `#include ` `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; ` `    ``// Intialise the ordered_set ` `    ``ordered_set set1; ` ` `  `    ``// Insert the first ` `    ``// element in set ` `    ``set1.insert(arr); ` ` `  `    ``// Intialise 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) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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