如何优化我的python代码下面给出的？

Question

I have an array and given an array of size N containing positive integers and I want to count number of smaller elements on right side of each array.

for example:-

    Input:
    N = 7
    arr[] = {12, 1, 2, 3, 0, 11, 4}
    Output: 6 1 1 1 0 1 0
    Explanation: There are 6 elements right
    after 12. There are 1 element right after
    1. And so on.
    

And my code for this problem is like as :- 
# python code here
    n=int(input())
    arr=list(map(int,input().split()))

    ans=0
    ANS=[]
    for i in range(n-1):
        for j in range(i+1,n):
            if arr[i]>arr[j]:
                ans+=1
        ANS.append(ans)
        ans=0
    ANS.append(0)
    print(ANS)

but the above my code take O(n^2) time complexity and I want to reduce the this. If anyone have any idea to reduce above python code time complexity please help me. Thank you.

Answer 1

This solution is O(n log(n)) as it is three iterations over the values and one sorting.

arr = [12, 1, 2, 3, 0, 11, 4]

# Gather original index and values
tups = []
for origin_index, el in enumerate(arr):
    tups.append([origin_index, el])

# sort on value
tups.sort(key=lambda t: t[1])

res = []
for sorted_index, values in enumerate(tups):
    # check the difference between the sorted and original index
    # If there is a positive value we have the n difference smaller 
    # values to the right of this index value. 
    if sorted_index - values[0] > 0:
        res.append([values[0], (sorted_index - values[0])])
    elif sorted_index - values[0] == 0:
        res.append([values[0], (sorted_index - values[0]) + 1])
    else:
        res.append([values[0], 0])

origin_sort_res = [0 for i in range(len(arr))]
for v in res:
    # Return the from the sorted array to the original indexing
    origin_sort_res[v[0]] = v[1]

print(origin_sort_res)

Answer 2

try this(nlog2n)

def solution(nums):
    sortns = []
    res = []
    for n in reversed(nums):
        idx = bisect.bisect_left(sortns, n)
        res.append(idx)
        sortns.insert(idx,n)
    return res[::-1]

print(solution([12, 1, 2, 3, 0, 11, 4]))
# [6, 1, 1, 1, 0, 1, 0]