Algorithm Question By Google

QUESTION

We can determine how "out of order" an array A is by counting the number of inversions it has.

Two elements A[i] and A[j] form an inversion if A[i] > A[j] but i < j.
That is, a smaller element appears after a larger element.

CHALLENGE

Given an array, count the number of inversions it has. Do this faster than O(N^2) time.

You may assume each element in the array is distinct.

For example, a sorted list has zero inversions.
The array [2, 4, 1, 3, 5] has three inversions: (2, 1), (4, 1), and (4, 3).
The array [5, 4, 3, 2, 1] has ten inversions: every distinct pair forms an inversion.

Comments

[Areahints]

Hi @theghostyced . so I decided to try and solve this on my own and I want to share my insights

my first solution was to write some logic google the question and I came across a solution that didn't satisfy the complexity

def getInvCount(arr, n): 

    inv_count = 0
    for i in range(n): 
        for j in range(i + 1, n): 
            if (arr[i] > arr[j]): 
                inv_count += 1

    return inv_count 

I then tried to rewrite this using list comprehension, you can see my effort here

def getInvCount(arr, n):
    return sum([1 if (arr[i] > arr[j]) else 0 for i in range(n) for j in range(i+1,n)]) 

but this also revealed the answers which you can use to solve this question.

[CED]

Thanks @areahints but I think the time complexity still remains the same irrespective of the list comprehension.

[Areahints]

yes, I mentioned that my list comprehension didn't satisfy the complexity.

consider this solution;

def getInvCount(arr,n):
    return mergeGetInvCount(arr)[1]

def mergeGetInvCount(arr):
    if len(arr) <= 1:
        return arr, 0
    middle = int(len(arr) / 2)
    left, a = mergeGetInvCount(arr[:middle])
    right, b = mergeGetInvCount(arr[middle:])
    result, c = mergeGetSplitInvCount(left, right)
    return result, (a + b + c)

def mergeGetSplitInvCount(left, right):
    result = []
    count = 0
    i,j = 0,0
    left_len = len(left)
    while i < left_len and j < len(right):
        if left[i] <= right[j]:
            result.append(left[i])
            i += 1
        else:
            result.append(right[j])
            count += left_len - i
            j += 1
    result += left[i:]
    result += right[j:]
    return result, count   

# Driver Code
arr = [1, 20, 6, 4, 5] 
n = len(arr) 
print("Number of inversions are", getInvCount(arr,n))

Output

Number of inversions are 5

that's an O(n log n) implementation that I mentioned has been revealed in the link. did you read through it?