Using Python, another approach to solve the above problem would be to use a simple Binary Search Tree with 2 extra fields:
- to hold the elements on the left side of a node
- 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)