Count smaller elements on right side and greater elements on left side using Binary Index Tree

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

Prerequisite: Counting inversions in an array using BIT

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:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// 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[]
int getSum(int BITree[], int index)
{
    int sum = 0; // Initialize result
  
    // Traverse ancestors of BITree[index]
    while (index > 0) {
        // Add current element of BITree to sum
        sum += BITree[index];
  
        // Move index to parent node in getSum View
        index -= index & (-index);
    }
    return sum;
}
  
// 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.
void updateBIT(int BITree[], int n, int index, int 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 }
void convert(int arr[], int n)
{
    // Create a copy of arrp[] in temp and sort the temp array
    // in increasing order
    int temp[n];
    for (int i = 0; i < n; i++)
        temp[i] = arr[i];
    sort(temp, temp + n);
  
    // Traverse all array elements
    for (int i = 0; i < n; i++) {
        // lower_bound() Returns pointer to the first element
        // greater than or equal to arr[i]
        arr[i] = lower_bound(temp, temp + n, arr[i]) - temp + 1;
    }
}
  
// Function to find smaller_right array
void findElements(int arr[], int 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)
    int BIT[n + 1];
    for (int i = 1; i <= n; i++)
        BIT[i] = 0;
  
    // To store smaller elements in right side
    // and greater elements on left side
    int smaller_right[n], greater_left[n];
  
    // Traverse all elements from right.
    for (int i = n - 1; i >= 0; i--) {
  
        // 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);
    }
  
    cout << "Smaller right: ";
  
    // Print smaller_right array
    for (int i = 0; i < n; i++)
        cout << smaller_right[i] << " ";
    cout << endl;
  
    for (int i = 1; i <= n; i++)
        BIT[i] = 0;
  
    // Find all left side greater elements
    for (int i = 0; i < n; i++) {
  
        // 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);
    }
  
    cout << "Greater left: ";
  
    // Print greater_left array
    for (int i = 0; i < n; i++)
        cout << greater_left[i] << " ";
}
  
// Driver code
int main()
{
    int arr[] = { 12, 1, 2, 3, 0, 11, 4 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function call
    findElements(arr, n);
  
    return 0;
}

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Python3

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# 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)
  
# This code is contributed by
# sanjeev2552

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Output:

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


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Improved By : sanjeev2552