## DEV Community

QUESTION

We can determine how "out of order" an array A is by counting the number of inversions it has.

Two elements A[i] and A[j] form an inversion if A[i] > A[j] but i < j.
That is, a smaller element appears after a larger element.

CHALLENGE

Given an array, count the number of inversions it has. Do this faster than O(N^2) time.

You may assume each element in the array is distinct.

For example, a sorted list has zero inversions.
The array [2, 4, 1, 3, 5] has three inversions: (2, 1), (4, 1), and (4, 3).
The array [5, 4, 3, 2, 1] has ten inversions: every distinct pair forms an inversion. Areahints

Hi @theghostyced . so I decided to try and solve this on my own and I want to share my insights

my first solution was to write some logic google the question and I came across a solution that didn't satisfy the complexity

``````def getInvCount(arr, n):

inv_count = 0
for i in range(n):
for j in range(i + 1, n):
if (arr[i] > arr[j]):
inv_count += 1

return inv_count
``````

I then tried to rewrite this using list comprehension, you can see my effort here

``````def getInvCount(arr, n):
return sum([1 if (arr[i] > arr[j]) else 0 for i in range(n) for j in range(i+1,n)])
``````

but this also revealed the answers which you can use to solve this question. CED

Thanks @areahints but I think the time complexity still remains the same irrespective of the list comprehension. Areahints

yes, I mentioned that my list comprehension didn't satisfy the complexity.

consider this solution;

``````def getInvCount(arr,n):
return mergeGetInvCount(arr)

def mergeGetInvCount(arr):
if len(arr) <= 1:
return arr, 0
middle = int(len(arr) / 2)
left, a = mergeGetInvCount(arr[:middle])
right, b = mergeGetInvCount(arr[middle:])
result, c = mergeGetSplitInvCount(left, right)
return result, (a + b + c)

def mergeGetSplitInvCount(left, right):
result = []
count = 0
i,j = 0,0
left_len = len(left)
while i < left_len and j < len(right):
if left[i] <= right[j]:
result.append(left[i])
i += 1
else:
result.append(right[j])
count += left_len - i
j += 1
result += left[i:]
result += right[j:]
return result, count

# Driver Code
arr = [1, 20, 6, 4, 5]
n = len(arr)
print("Number of inversions are", getInvCount(arr,n))
``````

## Output

``````Number of inversions are 5
``````

that's an O(n log n) implementation that I mentioned has been revealed in the link. did you read through it?