Good morning! Here's our prompt for today.

This was submitted by user Hackleman.

Determine how "out of order" a list is by writing a function that returns the number of inversions in that list.

Two elements of a list L, L[i] and L[j], form an inversion if L[i] < L[j] but i > j.

For example, the list [5, 4, 3, 2, 1] has 10 inversions since every element is out of order. They are the following: (5, 4), (5,3), (5,2), (5,1), (4,3), (4,2), (4,1), (3,2), (3,1), (2,1).

The list [2, 1, 4, 3, 5] has two inversions: (2, 1) and (4, 3).

Do this in better than O(N^2).

Constraints

• Length of the given array <= 100000
• Values in the array ranges from -1000000000 to 1000000000
• Expected time complexity : O(nlogn)
• Expected space complexity : O(1)

Try to solve this here or in Interactive Mode.

How do I practice this challenge?

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def inversions(l):

    # fill in

    return l

​

​

import unittest

​

​

class Test(unittest.TestCase):

    def test_1(self):

        assert inversions([5, 4, 3, 2, 1]) == 10

        print("PASSED: inversions for [5, 4, 3, 2, 1] should be 10")

​

    def test_2(self):

        assert inversions([2, 4, 1, 3, 5]) == 3

        print("PASSED: inversions for [2, 4, 1, 3, 5] should be 3")

​

    def test_3(self):

        assert callable(inversion) == True

        print("PASSED: `inversions` should be a function")

​

​

if __name__ == "__main__":

    unittest.main(verbosity=2)

    print("Nice job, 3/3 tests passed!")

​

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