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

06 Nov, 2021
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 2Input: 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++
Java
Python3
C#
// 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 arrayvoid 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 codeint main(){    int arr[] = { 12, 1, 2, 3, 0, 11, 4 };     int n = sizeof(arr) / sizeof(arr[0]);     // Function call    findElements(arr, n);     return 0;}
// Java implementation of the approachimport java.io.*;import java.util.*; class GFG{     // 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[]public static int getSum(int BITree[], int index){         // Initialize result    int sum = 0;     // 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.public static 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 }public static void convert(int arr[], int n){         // Create a copy of arrp[] in temp    // and sort the temp array in    // increasing order    int[] temp = new int[n];    for(int i = 0; i < n; i++)        temp[i] = arr[i];         Arrays.sort(temp);     // Traverse all array elements    for(int i = 0; i < n; i++)    {                 // Arrays.binarySearch() returns index        // to the first element greater than        // or equal to arr[i]        arr[i] = Arrays.binarySearch(temp, arr[i]) + 1;    }} // Function to find smaller_right arraypublic static 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 = new int[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 = new int[n];    int[] greater_left = new int[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);    }         System.out.print("Smaller right: ");     // Print smaller_right array    for(int i = 0; i < n; i++)        System.out.print(smaller_right[i] + " ");         System.out.println();     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);    }     System.out.print("Greater left: ");     // Print greater_left array    for(int i = 0; i < n; i++)        System.out.print(greater_left[i] + " ");} // Driver codepublic static void main(String[] args){    int arr[] = { 12, 1, 2, 3, 0, 11, 4 };     int n = arr.length;     // Function call    findElements(arr, n);}} // This code is contributed by kunalsg18elec   
# Python3 implementation of the approachfrom 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 arraydef 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 Codeif __name__ == "__main__":     arr = [12, 1, 2, 3, 0, 11, 4]    n = len(arr)     # Function call    findElements(arr, n) # This code is contributed by# sanjeev2552
// C# implementation of the// above approachusing System;class GFG{     // 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[]public static int getSum(int []BITree,                         int index){      // Initialize result  int sum = 0;   // 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.public static 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 }public static void convert(int []arr,                           int n){      // Create a copy of arrp[] in temp  // and sort the temp array in  // increasing order  int[] temp = new int[n];     for(int i = 0; i < n; i++)    temp[i] = arr[i];   Array.Sort(temp);   // Traverse all array elements  for(int i = 0; i < n; i++)  {    // Arrays.binarySearch()    // returns index to the    // first element greater    // than or equal to arr[i]    arr[i] =  Array.BinarySearch(temp,                                 arr[i]) + 1;  }} // Function to find smaller_right arraypublic static 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 = new int[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 = new int[n];  int[] greater_left = new int[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);  }   Console.Write("Smaller right: ");   // Print smaller_right array  for(int i = 0; i < n; i++)    Console.Write(smaller_right[i] + " ");   Console.WriteLine();   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);  }   Console.Write("Greater left: ");   // Print greater_left array  for(int i = 0; i < n; i++)    Console.Write(greater_left[i] + " ");} // Driver codepublic static void Main(String[] args){  int []arr = {12, 1, 2,               3, 0, 11, 4};  int n = arr.Length;   // Function call  findElements(arr, n);}}    // This code is contributed by Rajput-Ji
Smaller right: 6 1 1 1 0 1 0 
Greater left: 0 1 1 1 4 1 2
 
Time Complexity: O(N2)
Auxiliary Space: O(N)
sanjeev2552
kunalsg18elec
Rajput-Ji
rohan07
Binary Indexed Tree
Advanced Data Structure
Arrays
Bit Magic
Arrays
Bit Magic
Binary Indexed Tree
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