Algorithm Number of Inversions (Python)

Question

I'm attempting to complete a code assignment for the Algorithmic Toolbox offered by UC San Diego on Coursera. The assignment requires that you count the number of inversions in a sequence of numbers using a variation of the merge-sort algorithm. For a better description of the problem:

https://i.stack.imgur.com/CCBU8.jpg

I've used a variation of a merge-sort algorithm but am getting an incorrect answer and am frankly stuck.

Of note is that before explaining what I've attempted is that this code involves recursion which I admit I'm finding tricky to understand.

Mostly beyond the usual debugging I've tried to compare my code to known solutions to see where I may be going wrong. I could submit those as my solution but as far as I'm concerned that would be a cheat and I'd like to know where I've gone wrong with my code (and it's quite honestly driving me nuts).

#Uses python3
import sys

def merge_sort(A):
    if len(A) <= 1:
        return A, 0    
    else:
        middle = (len(A)//2)
        left, count_left = merge_sort(A[:middle])
        right, count_right = merge_sort(A[middle:])
        result, count_result = merge(left,right)
        return result, (count_left + count_right + count_result)

def merge(a,b):
    result = []
    count = 0
    while len(a) > 0 and len(b) > 0:
        if a[0] < b[0]:
            result.append(a[0])
            a.remove(a[0])
        else:
            #count = count + (len(a) - 1)
            result.append(b[0])
            b.remove(b[0])
            count += (len(a) - 1) #this is the important line
    if len(a) == 0:
        result = result + b
    else:
        result = result + a
    return result, count 

if __name__ == '__main__':
    input = sys.stdin.read()
    n, *a = list(map(int, input.split()))
    c = n * [0]
    array, inversion = merge_sort(a)
    print(array)
    print(inversion)

Listed below are two sample inputs I have been using in my testing:

# ex 1:
3
3 1 2

Note that the first digit is the number of values in the sequence (required for the grader). Expected answer for inversions is 2. I'm getting 0 with my code.

# ex 2:
6
3 1 9 3 2 1

Expected answer for inversions is 9. I'm getting 4.

MBo · Accepted Answer · 2019-02-19T15:12:45.987

Two corrections:

if a[0] <= b[0]: (note that a lot of internet examples and courses ignore or equal case, destroying intrinsic algorithm stability, this case also is important for correct inv. counting)

and count += len(a) - we need to account that all items in a form inversions with current b item

def merge(a,b):
    result = []
    count = 0
    while len(a) > 0 and len(b) > 0:
        if a[0] <= b[0]:   
            result.append(a[0])
            a.remove(a[0])
        else:
            result.append(b[0])
            b.remove(b[0])
            count += len(a)
    if len(a) == 0:
        result = result + b
    else:
        result = result + a
    return result, count

Thank you for the corrections and more importantly the explanations this seems to have done the trick. This particular small algorithm has had me stumped and humbled for good amount of time. — roadrunner007, Feb 20 '19 at 05:58

Algorithm Number of Inversions (Python)

1 Answers1