Binary Index Tree using Python3

Software Development
#1

Given an array arr[] of size N . The task is to find smaller elements on the right side and greater elements on the left side for each element arr[i] in the given array.
Examples:

Input: arr[] = {12, 1, 2, 3, 0, 11, 4}
Output:
Smaller right: 6 1 1 1 0 1 0
Greater left: 0 1 1 1 4 1 2
Input: arr[] = {5, 4, 3, 2, 1}
Output:
Smaller right: 4 3 2 1 0
Greater left: 0 1 2 3 4
Input: arr[] = {1, 2, 3, 4, 5}
Output:
Smaller right: 0 0 0 0 0
Greater left: 0 0 0 0 0

Approach: We have already discussed the implementation to count smaller elements on the right side in this post. Here, we will use Binary Indexed Tree to count smaller elements on the right side and greater elements on the left side for each element in the array. First, traverse the array from right to left and find smaller elements on the right side as suggested in the previous post. Then reset the BIT array and traverse the array from left to right and find greater elements on the left side.
Below is the implementation of the above approach:

  • Python3

# Python3 implementation of the approach

from bisect import bisect_left as lower_bound

# Function to return the sum of arr[0..index]

# This function assumes that the array is

# preprocessed and partial sums of array elements

# are stored in BITree[]

def getSum(BITree, index):

# Initialize result

s = 0

# Traverse ancestors of BITree[index]

while index > 0 :

# Add current element of BITree to sum

s + = BITree[index]

# Move index to parent node in getSum View

index - = index & ( - index)

return s

# Updates a node in Binary Index Tree (BITree)

# at given index in BITree. The given value 'val'

# is added to BITree[i] and all of its ancestors in tree.

def updateBIT(BITree, n, index, val):

# Traverse all ancestors and add 'val'

while index < = n:

# Add 'val' to current node of BI Tree

BITree[index] + = val

# Update index to that of parent in update View

index + = index & ( - index)

# Converts an array to an array with values

# from 1 to n and relative order of smaller

# and greater elements remains same.

# For example, {7, -90, 100, 1} is

# converted to {3, 1, 4, 2 }

def convert(arr, n):

# Create a copy of arrp[] in temp and

# sort the temp array in increasing order

temp = [ 0 ] * n

for i in range (n):

temp[i] = arr[i]

temp.sort()

# Traverse all array elements

for i in range (n):

# lower_bound() Returns pointer to the first element

# greater than or equal to arr[i]

arr[i] = lower_bound(temp, arr[i]) + 1

# Function to find smaller_right array

def findElements(arr, n):

# Convert arr[] to an array with values

# from 1 to n and relative order of smaller and

# greater elements remains same. For example,

# {7, -90, 100, 1} is converted to {3, 1, 4, 2 }

convert(arr, n)

# Create a BIT with size equal to maxElement+1

# (Extra one is used so that elements can be

# directly be used as index)

BIT = [ 0 ] * (n + 1 )

# To store smaller elements in right side

# and greater elements on left side

smaller_right = [ 0 ] * n

greater_left = [ 0 ] * n

# Traverse all elements from right.

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

# Get count of elements smaller than arr[i]

smaller_right[i] = getSum(BIT, arr[i] - 1 )

# Add current element to BIT

updateBIT(BIT, n, arr[i], 1 )

print ( "Smaller right:" , end = " " )

for i in range (n):

print (smaller_right[i], end = " " )

print ()

# Print smaller_right array

for i in range ( 1 , n + 1 ):

BIT[i] = 0

# Find all left side greater elements

for i in range (n):

# Get count of elements greater than arr[i]

greater_left[i] = i - getSum(BIT, arr[i])

# Add current element to BIT

updateBIT(BIT, n, arr[i], 1 )

print ( "Greater left:" , end = " " )

# Print greater_left array

for i in range (n):

print (greater_left[i], end = " " )

print ()

# Driver Code

if __name__ = = "__main__" :

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

n = len (arr)

# Function call

findElements(arr, n)

Output:

Smaller right: 6 1 1 1 0 1 0
Greater left: 0 1 1 1 4 1 2