# Count inversions in an array | Set 4 ( 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.

We have already discussed below methods to solve inversion count –

1. Naive and Modified Merge Sort
2. Using AVL Tree
3. 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++ implementation ` `#include ` `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; ` `} `

Output:

```Number of inversions are : 6
```

Time Complexity: Auxiliary Space: My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

2

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.