Add count_inversions.py

Author: kumailn

Python/Count_Inversions.py

@@ -0,0 +1,37 @@
+def countInversions(nums):
+    #Our recursive base case, if our list is of size 1 we know there's no inversion and that it's already sorted
+    if len(nums) == 1: return nums, 0
+
+    #We run our function recursively on it's left and right halves
+    left, leftInversions = countInversions(nums[:len(nums) // 2])
+    right, rightInversions = countInversions(nums[len(nums) // 2:])
+
+    #Initialize our inversions variable to be the number of inversions in each half
+    inversions = leftInversions + rightInversions
+
+    #Initialize a list to hold our sorted result list
+    sortedNums = []
+
+    i = j = 0
+    #Here we count the inversions that exist between the two halves -- while sorting them at the same time
+    while i < len(left) and j < len(right):
+        if left[i] < right[j]:
+            sortedNums += [left[i]]; i += 1
+        else:
+            sortedNums += [right[j]]; j += 1
+
+            #Since we know that 'left' is sorted, once we reach an item in 'left' thats bigger than something in right that item and everything to it's right-hand side must be an inversion!
+            #This line right here is exactly what shrinks the time of this algorithm from n^2 to nlogn as it means we don't need to compare every single pair
+            inversions += len(left) - i
+
+    #Once we've exhausted either left or right, we can just add the other one onto our sortedNums list
+    sortedNums += left[i:] + right[j:]
+
+    return sortedNums, inversions
+
+
+def main():
+    nums = [1, 5, 4, 8, 10, 2, 6, 9]
+    print(countInversions(nums))
+
+main()