given an array, for each element, find out the total number of elements lesser than it, which appear to the right of it

Question

I had previously posted a question, Given an array, find out the next smaller element for each element now, i was trying to know , if there is any way to find out \"given an ar

Answer 1

You can also use an array instead of a binary search tree.

def count_next_smaller_elements(xs):
    # prepare list "ys" containing item's numeric order
    ys = sorted((x,i) for i,x in enumerate(xs))
    zs = [0] * len(ys)

    for i in range(1, len(ys)):
        zs[ys[i][1]] = zs[ys[i-1][1]]
        if ys[i][0] != ys[i-1][0]: zs[ys[i][1]] += 1

    # use list "ts" as binary search tree, every element keeps count of
    # number of children with value less than the current element's value
    ts = [0] * (zs[ys[-1][1]]+1)
    us = [0] * len(xs)

    for i in range(len(xs)-1,-1,-1):
        x = zs[i]+1
        while True:
            us[i] += ts[x-1]
            x -= (x & (-x))
            if x <= 0: break

        x = zs[i]+1
        while True:
            x += (x & (-x))
            if x > len(ts): break
            ts[x-1] += 1

    return us

print count_next_smaller_elements([40, 20, 10, 50, 20, 40, 30])
# outputs: [4, 1, 0, 2, 0, 1, 0]