Counting inversions in an array

Question

I\'m designing an algorithm to do the following: Given array A[1... n], for every i < j, find all inversion pairs such that A[i] > A[j]

Answer 1

Solution One.Work well when there is a large amount of numbers

def countInversions(arr):
 n = len(arr)
 if n == 1:
    return 0
 n1 = n // 2
 n2 = n - n1
 arr1 = arr[:n1]
 arr2 = arr[n1:]
 # print(n1,'||',n1,'||',arr1,'||',arr2)
 ans = countInversions(arr1) + countInversions(arr2)
 print(ans)
 i1 = 0
 i2 = 0
 for i in range(n):
     # print(i1,n1,i2,n2)
     if i1 < n1 and (i2 >= n2 or arr1[i1] <= arr2[i2]):
         arr[i] = arr1[i1]
         ans += i2
         i1 += 1
     elif i2 < n2:
         arr[i] = arr2[i2]
         i2 += 1
 return ans

Solution Two.Simple solution.

def countInversions(arr):
      count = 0
      for i in range(len(arr)):
          for j in range(i, len(arr)):
               # print(arr[i:len(arr)])
                 if arr[i] > arr[j]:
                     print(arr[i], arr[j])
                     count += 1
      print(count)