Approach:
Suppose the number of inversions in the left half and right half of the array (let be inv1 and inv2); what kinds of inversions are not accounted for in Inv1 + Inv2? The answer is – the inversions that need to be counted during the merge step. Therefore, to get the total number of inversions that needs to be added are the number of inversions in the left subarray, right subarray, and merge().
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How to get the number of inversions in merge()?
In merge process, let i is used for indexing left sub-array and j for right sub-array. At any step in merge(), if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j] -
Algorithm:
- The idea is similar to merge sort, divide the array into two equal or almost equal halves in each step until the base case is reached.
- Create a function merge that counts the number of inversions when two halves of the array are merged, create two indices i and j, i is the index for the first half, and j is an index of the second half. if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j].
- Create a recursive function to divide the array into halves and find the answer by summing the number of inversions is the first half, the number of inversion in the second half and the number of inversions by merging the two.
- The base case of recursion is when there is only one element in the given half.
- Print the answer
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Implementation:
# Python 3 program to count inversions in an array
# Function to Use Inversion Count
def mergeSort(arr, n):
# A temp_arr is created to store
# sorted array in merge function
temp_arr = [ 0 ] * n
return _mergeSort(arr, temp_arr, 0 , n - 1 )
# This Function will use MergeSort to count inversions
def _mergeSort(arr, temp_arr, left, right):
# A variable inv_count is used to store
# inversion counts in each recursive call
inv_count = 0
# We will make a recursive call if and only if
# we have more than one elements
if left < right:
# mid is calculated to divide the array into two subarrays
# Floor division is must in case of python
mid = (left + right) / / 2
# It will calculate inversion
# counts in the left subarray
inv_count + = _mergeSort(arr, temp_arr,
left, mid)
# It will calculate inversion
# counts in right subarray
inv_count + = _mergeSort(arr, temp_arr,
mid + 1 , right)
# It will merge two subarrays in
# a sorted subarray
inv_count + = merge(arr, temp_arr, left, mid, right)
return inv_count
# This function will merge two subarrays
# in a single sorted subarray
def merge(arr, temp_arr, left, mid, right):
i = left # Starting index of left subarray
j = mid + 1 # Starting index of right subarray
k = left # Starting index of to be sorted subarray
inv_count = 0
# Conditions are checked to make sure that
# i and j don't exceed their
# subarray limits.
while i < = mid and j < = right:
# There will be no inversion if arr[i] <= arr[j]
if arr[i] < = arr[j]:
temp_arr[k] = arr[i]
k + = 1
i + = 1
else :
# Inversion will occur.
temp_arr[k] = arr[j]
inv_count + = (mid - i + 1 )
k + = 1
j + = 1
# Copy the remaining elements of left
# subarray into temporary array
while i < = mid:
temp_arr[k] = arr[i]
k + = 1
i + = 1
# Copy the remaining elements of right
# subarray into temporary array
while j < = right:
temp_arr[k] = arr[j]
k + = 1
j + = 1
# Copy the sorted subarray into Original array
for loop_var in range (left, right + 1 ):
arr[loop_var] = temp_arr[loop_var]
return inv_count
# Driver Code
# Given array is
arr = [ 1 , 20 , 6 , 4 , 5 ]
n = len (arr)
result = mergeSort(arr, n)
print ( "Number of inversions are" , result)
Output:
Number of inversions are 5
Complexity Analysis:
- Time Complexity: O(n log n), The algorithm used is divide and conquer, So in each level, one full array traversal is needed, and there are log n levels, so the time complexity is O(n log n).
- Space Complexity**:** O(n), Temporary array.