Using BST with 2 extra fields using Python

Using Python, another approach to solve the above problem would be to use a simple Binary Search Tree with 2 extra fields:

  1. to hold the elements on the left side of a node
  2. to store the frequency of element.

In this approach, we traverse the input array from the ending to the begging and add the elements into the BST.
While inserting the elements to the BST, we can compute the number of elements which are lesser elements simply by computing the sum of frequency of the element and the number of elements to the left side of current node, if we are moving to right side of the current node.

Once we place an element in it’s correct position, we can return it’s this sum value

  • C++14
  • Python

class Node:

def __init__( self ,val):

self .val = val

self .left = None

self .right = None

# denotes number of times (frequency)

# an element has occurred.

self .elecount = 1

# denotes the number of nodes on left

# side of the node encountered so far.

self .lcount = 0

class Tree:

def __init__( self ,root):

self .root = root

def insert( self ,node):

"""This function helps to place an element at

its correct position in the BST and returns

the count of elements which are smaller than

the elements which are already inserted into the BST.

"""

curr = self .root

cnt = 0

while curr! = None :

prev = curr

if node.val>curr.val:

# This step computes the number of elements

# which are less than the current Node.

cnt + = (curr.elecount + curr.lcount)

curr = curr.right

elif node.val<curr.val:

curr.lcount + = 1

curr = curr.left

else :

prev = curr

prev.elecount + = 1

break

if prev.val>node.val:

prev.left = node

elif prev.val<node.val:

prev.right = node

else :

return cnt + prev.lcount

return cnt

def constructArray(arr,n):

t = Tree(Node(arr[ - 1 ]))

ans = [ 0 ]

for i in range (n - 2 , - 1 , - 1 ):

ans.append(t.insert(Node(arr[i])))

return reversed (ans)

# Driver function for above code

def main():

n = 7

arr = [ 10 , 6 , 15 , 20 , 30 , 5 , 7 ]

print ( " " .join( list ( map ( str ,constructArray(arr,n)))))

if __name__ = = "__main__" :

main()

# Code Contributed by Brahmajit Mohapatra

Output:

3 1 2 2 2 0 0

Time Complexity: O(nLogn)
Auxiliary Space: O(n)