# Count Inversions - ll

Posted: 6 Oct, 2020
Difficulty: Moderate

## PROBLEM STATEMENT

#### Two nodes 'N1' and 'N2' form an inversion if the value of 'N1' is greater than the value of 'N2' and 'N1' appears before 'N2' in the linked list.

##### Input Format:
``````The first and the only line of the input contains the elements of the singly linked list separated by a single space and terminated by -1. Hence, -1 would not be a list element.
``````
##### Output Format:
``````Print a single integer denoting the inversion count of the given linked list.
``````
##### Note:
``````You don't need to print the output, it has already been taken care of. Just implement the given function.
``````
##### Constraints:
``````0 <= L <= 10^5
-10^9 <= data <= 10^9 and data != -1

Where L is the number of nodes in the linked list and 'data' is the node value of the linked list.

Time Limit: 1 sec
`````` Approach 1

Initially, we declare a counter ‘INVERSIONCNT’ and initialize it with ‘0’, Where  ‘INVERSIONCNT’ maintains the count of inversions in the given linked list. Also, Declare a pointer ‘CUR’, initialized to the ‘HEAD’ of the linked list, to traverse the entire list, and also a pointer ‘TEMP’ to traverse the part of the linked list after ‘CUR’.

For every node pointed by the ‘CUR’, we use the ‘TEMP’ pointer to start traversal from the node next to the ‘CUR’. We compare the values of the nodes pointed by the pointer ‘TEMP’ and the ‘CUR’ pointer and if the value pointed by ‘CUR’ is greater than that pointed by ‘TEMP’, then the 2 nodes form an inversion pair and we increment the counter ‘INVERSIONCNT’ by ‘1’.

We repeat the entire process till ‘CUR’ reaches the end of the list.