re: Algorithm Question By Google VIEW POST

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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)[1]

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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