Software Engineer

The idea is similar to merge sort, divide the array into two equal or almost equal halves in each step until the base case is reached.
Create a function merge that counts the number of inversions when two halves of the array are merged, create two indices i and j, i is the index for the first half, and j is an index of the second half. if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j].
Create a recursive function to divide the array into halves and find the answer by summing the number of inversions is the first half, the number of inversion in the second half and the number of inversions by merging the two.
The base case of recursion is when there is only one element in the given half.

Have a visited flag with each node.
Traverse the linked list and keep marking visited nodes.
If you see a visited node again then there is a loop. This solution works in O(n) but requires additional information with each node.
A variation of this solution that doesn't require modification to basic data structure can be implemented using a hash, just store the addresses of visited nodes in a hash and if you see an address that already exists in hash then there is a loop.

Used Greedy approach.