### re: Algorithm Question By Google VIEW POST

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

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?

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