Count smaller elements on Right side
Last Updated : 23 May, 2024
Given an unsorted array arr[] of distinct integers, construct another array countSmaller[] such that countSmaller[i] contains the count of smaller elements on the right side of each element arr[i] in the array.
Examples:
Input: arr[] = {12, 1, 2, 3, 0, 11, 4}
Output: countSmaller[] = {6, 1, 1, 1, 0, 1, 0}
Input: arr[] = {5, 4, 3, 2, 1}
Output: countSmaller[] = {4, 3, 2, 1, 0}
Input: arr[] = {1, 2, 3, 4, 5}
Output: countSmaller[] = {0, 0, 0, 0, 0}
Naive Approach:
Use nested loops. The outer loop picks all elements from left to right. The inner loop iterates through all the elements on right side of the picked element and updates countSmaller[].
Below is the implementation of the above idea.
C++ #include <iostream> using namespace std; void constructLowerArray(int arr[], int* countSmaller, int n) { int i, j; // Initialize all the counts in // countSmaller array as 0 for (i = 0; i < n; i++) countSmaller[i] = 0; for (i = 0; i < n; i++) { for (j = i + 1; j < n; j++) { if (arr[j] < arr[i]) countSmaller[i]++; } } } // Utility function that prints // out an array on a line void printArray(int arr[], int size) { int i; for (i = 0; i < size; i++) cout << arr[i] << " "; cout << "\n"; } // Driver code int main() { int arr[] = { 12, 1, 2, 3, 0, 11, 4 }; int n = sizeof(arr) / sizeof(arr[0]); int* low = (int*)malloc(sizeof(int) * n); constructLowerArray(arr, low, n); printArray(low, n); return 0; } // This code is contributed by Hemant Jain void constructLowerArray(int arr[], int* countSmaller, int n) { int i, j; // initialize all the counts in countSmaller array as 0 for (i = 0; i < n; i++) countSmaller[i] = 0; for (i = 0; i < n; i++) { for (j = i + 1; j < n; j++) { if (arr[j] < arr[i]) countSmaller[i]++; } } } /* Utility function that prints out an array on a line */ void printArray(int arr[], int size) { int i; for (i = 0; i < size; i++) printf("%d ", arr[i]); printf("\n"); } // Driver program to test above functions int main() { int arr[] = { 12, 1, 2, 3, 0, 11, 4 }; int n = sizeof(arr) / sizeof(arr[0]); int* low = (int*)malloc(sizeof(int) * n); constructLowerArray(arr, low, n); printArray(low, n); return 0; } class CountSmaller { void constructLowerArray(int arr[], int countSmaller[], int n) { int i, j; // initialize all the counts in countSmaller array // as 0 for (i = 0; i < n; i++) countSmaller[i] = 0; for (i = 0; i < n; i++) { for (j = i + 1; j < n; j++) { if (arr[j] < arr[i]) countSmaller[i]++; } } } /* Utility function that prints out an array on a line */ void printArray(int arr[], int size) { int i; for (i = 0; i < size; i++) System.out.print(arr[i] + " "); System.out.println(""); } // Driver program to test above functions public static void main(String[] args) { CountSmaller small = new CountSmaller(); int arr[] = { 12, 1, 2, 3, 0, 11, 4 }; int n = arr.length; int low[] = new int[n]; small.constructLowerArray(arr, low, n); small.printArray(low, n); } } def constructLowerArray(arr, countSmaller, n): # initialize all the counts in countSmaller array as 0 for i in range(n): countSmaller[i] = 0 for i in range(n): for j in range(i + 1, n): if (arr[j] < arr[i]): countSmaller[i] += 1 # Utility function that prints out an array on a line def printArray(arr, size): for i in range(size): print(arr[i], end=" ") print() # Driver code arr = [12, 1, 2, 3, 0, 11, 4] n = len(arr) low = [0]*n constructLowerArray(arr, low, n) printArray(low, n) # This code is contributed by ApurvaRaj
using System; class GFG { static void constructLowerArray(int[] arr, int[] countSmaller, int n) { int i, j; // initialize all the counts in // countSmaller array as 0 for (i = 0; i < n; i++) countSmaller[i] = 0; for (i = 0; i < n; i++) { for (j = i + 1; j < n; j++) { if (arr[j] < arr[i]) countSmaller[i]++; } } } /* Utility function that prints out an array on a line */ static void printArray(int[] arr, int size) { int i; for (i = 0; i < size; i++) Console.Write(arr[i] + " "); Console.WriteLine(""); } // Driver function public static void Main() { int[] arr = new int[] { 12, 1, 2, 3, 0, 11, 4 }; int n = arr.Length; int[] low = new int[n]; constructLowerArray(arr, low, n); printArray(low, n); } } // This code is contributed by Sam007 function constructLowerArray(arr, countSmaller, n) { let i, j; // initialize all the counts in // countSmaller array as 0 for (i = 0; i < n; i++) countSmaller[i] = 0; for (i = 0; i < n; i++) { for (j = i + 1; j < n; j++) { if (arr[j] < arr[i]) countSmaller[i]++; } } } /* Utility function that prints out an array on a line */ function printArray(arr, size) { let i; for (i = 0; i < size; i++) console.log(arr[i] + " "); } let arr = [12, 1, 2, 3, 0, 11, 4]; let n = arr.length; let low = new Array(n); constructLowerArray(arr, low, n); printArray(low, n); Time Complexity: O(N2), Used nested loop for calculating every element in the array of size N.
Auxiliary Space: O(N)
Count smaller elements on the right side using Merge Sort:
The idea is to divide the array into two halves just as we do in merge sort. And then while merging back we sort them in decreasing order and keep track of count the smaller elements.
Follow the steps below to solve the problem:
- Declare an array of pair of int V and an array ans to keep track of count. Traverse the given array and insert all the elements and their corresponding index as a pair in this new array. We will call our merge sort function on this array.
- In the merge function, we declare two pointers i and j as usual to traverse our left and right array and merge them in descending order.
- Now, as we traverse these two arrays, We know that all the elements that are present in right array are also on the right side (in the actual array) of the elements that are present in left array.
- So, whenever we find a element in left array which is greater than the element in the right array( V[i].first > V[j].first ) we increase our count ans[ V[i].second ] += size of right array - j +1. Because our arrays are in decreasing order that means all the elements next to it are also smaller than V[i].first.
- After increasing the count we just simply do our sorting in decreasing and move our pointers ahead.
Below is the implementation of the above approach:
C++ #include <bits/stdc++.h> using namespace std; void merge(vector<pair<int, int> >& v, vector<int>& ans, int l, int mid, int h) { vector<pair<int, int> > t; // temporary array for merging both halves int i = l; int j = mid + 1; while (i < mid + 1 && j <= h) { // v[i].first is greater than all // the elements from j till h. if (v[i].first > v[j].first) { ans[v[i].second] += (h - j + 1); t.push_back(v[i]); i++; } else { t.push_back(v[j]); j++; } } // if any elements left in left array while (i <= mid) t.push_back(v[i++]); // if any elements left in right array while (j <= h) t.push_back(v[j++]); // putting elements back in main array in // descending order for (int k = 0, i = l; i <= h; i++, k++) v[i] = t[k]; } void mergesort(vector<pair<int, int> >& v, vector<int>& ans, int i, int j) { if (i < j) { int mid = (i + j) / 2; // calling mergesort for left half mergesort(v, ans, i, mid); // calling mergesort for right half mergesort(v, ans, mid + 1, j); // merging both halves and generating answer merge(v, ans, i, mid, j); } } vector<int> constructLowerArray(int* arr, int n) { vector<pair<int, int> > v; // inserting elements and corresponding index // as pair for (int i = 0; i < n; i++) v.push_back({ arr[i], i }); // answer array for keeping count // initialized by 0, vector<int> ans(n, 0); // calling mergesort mergesort(v, ans, 0, n - 1); return ans; } // Driver Code Starts. int main() { int arr[] = { 12, 1, 2, 3, 0, 11, 4 }; int n = sizeof(arr) / sizeof(arr[0]); auto ans = constructLowerArray(arr, n); for (auto x : ans) { cout << x << " "; } cout << "\n"; return 0; } // Driver code ends. // This code is contributed by Manjeet Singh import java.util.*; class LowerArray { static void merge(int[][] v, int[] ans, int l, int mid, int h) { int[][] t = new int[h - l + 1][2]; // temporary array for merging both halves int i = l; int j = mid + 1; int k = 0; while (i < mid + 1 && j <= h) { // v[i][0] is greater than all // the elements from j till h. if (v[i][0] > v[j][0]) { ans[v[i][1]] += (h - j + 1); t[k][0] = v[i][0]; t[k][1] = v[i][1]; i++; } else { t[k][0] = v[j][0]; t[k][1] = v[j][1]; j++; } k++; } // if any elements left in left array while (i <= mid) { t[k][0] = v[i][0]; t[k][1] = v[i][1]; k++; i++; } // if any elements left in right array while (j <= h) { t[k][0] = v[j][0]; t[k][1] = v[j][1]; k++; j++; } // putting elements back in main array in // descending order k = 0; for (i = l; i <= h; i++, k++) { v[i][0] = t[k][0]; v[i][1] = t[k][1]; } } static void mergesort(int[][] v, int[] ans, int i, int j) { if (i < j) { int mid = (i + j) / 2; // calling mergesort for left half mergesort(v, ans, i, mid); // calling mergesort for right half mergesort(v, ans, mid + 1, j); // merging both halves and generating answer merge(v, ans, i, mid, j); } } static int[] constructLowerArray(int[] arr, int n) { int[][] v = new int[n][2]; // inserting elements and corresponding index // as pair for (int i = 0; i < n; i++) { v[i][0] = arr[i]; v[i][1] = i; } // answer array for keeping count // initialized by 0, int[] ans = new int[n]; // calling mergesort mergesort(v, ans, 0, n - 1); return ans; } // Driver Code Starts. public static void main(String[] args) { int[] arr = { 12, 1, 2, 3, 0, 11, 4 }; int n = arr.length; int[] ans = constructLowerArray(arr, n); for (int x : ans) { System.out.print(x + " "); } System.out.println(); } } def merge(v, ans, l, mid, h): t = [] # temporary array for merging both halves i = l j = mid+1 while (i < mid+1 and j <= h): # v[i][0] is greater than all # the elements from j till h. if v[i][0] > v[j][0]: ans[v[i][1]] += (h-j+1) t.append(v[i]) i += 1 else: t.append(v[j]) j += 1 # if any elements left in left array while (i <= mid): t.append(v[i]) i += 1 # if any elements left in right array while j <= h: t.append(v[j]) j += 1 # putting elements back in main array in # descending order k = 0 i = l while (i <= h): v[i] = t[k] i += 1 k += 1 def mergesort(v, ans, i, j): if i < j: mid = (i+j)//2 # calling mergesort for left half mergesort(v, ans, i, mid) # calling mergesort for right half mergesort(v, ans, mid + 1, j) # merging both halves and generating answer merge(v, ans, i, mid, j) def constructLowerArray(arr, n): v = [] # inserting elements and corresponding index as pair for i in range(n): v.append([arr[i], i]) # answer array for keeping count initialized by 0 ans = [0]*n # calling mergesort mergesort(v, ans, 0, n-1) return ans # Driver Code arr = [12, 1, 2, 3, 0, 11, 4] n = len(arr) ans = constructLowerArray(arr, n) for x in ans: print(x, end=" ") '''This code is contributed by RAJATKUMARGLA19'''
using System; using System.Collections.Generic; class Program { static void Merge(List<Tuple<int, int>> v, int[] ans, int l, int mid, int h) { List<Tuple<int, int>> t = new List<Tuple<int, int>>();// temporary array for merging both halves int i = l; int j = mid + 1; while (i < mid + 1 && j <= h) { // v[i].first is greater than all // the elements from j till h. if (v[i].Item1 > v[j].Item1) { ans[v[i].Item2] += (h - j + 1); t.Add(v[i]); i++; } else { t.Add(v[j]); j++; } } // if any elements left in left array while (i <= mid) t.Add(v[i++]); // if any elements left in right array while (j <= h) t.Add(v[j++]); i =l; // putting elements back in main array in // descending order for (int k = 0; i <= h; i++, k++) v[i] = t[k]; } static void MergeSort(List<Tuple<int, int>> v, int[] ans, int i, int j) { if (i < j) { int mid = (i + j) / 2; // calling mergesort for left half MergeSort(v, ans, i, mid); // calling mergesort for right half MergeSort(v, ans, mid + 1, j); // merging both halves and generating answer Merge(v, ans, i, mid, j); } } static int[] ConstructLowerArray(int[] arr) { List<Tuple<int, int>> v = new List<Tuple<int, int>>(); // inserting elements and corresponding index // as pair for (int i = 0; i < arr.Length; i++) v.Add(new Tuple<int, int>(arr[i], i)); // answer array for keeping count // initialized by 0, int[] ans = new int[arr.Length]; // calling mergesort MergeSort(v, ans, 0, arr.Length - 1); return ans; } // Driver code static void Main(string[] args) { int[] arr = { 12, 1, 2, 3, 0, 11, 4 }; int[] ans = ConstructLowerArray(arr); foreach (int x in ans) { Console.Write(x + " "); } Console.WriteLine(); } } // This code is contributed by ratiagrawal. function merge( v,ans, l, mid, h) { let t=[]; // temporary array for merging both halves let i = l; let j = mid + 1; while (i < mid + 1 && j <= h) { // v[i].first is greater than all // the elements from j till h. if (j < v.length && v[i][0] > v[j][0]) { ans[v[i][1]] += (h - j + 1); t.push(v[i]); i++; } else { t.push(v[j]); j++; } } // if any elements left in left array while (i <= mid) { t.push(v[i]); i++; } // if any elements left in right array while (j <= h) { t.push(v[j]); j++; } // putting elements back in main array in // descending order for (let k = 0, i = l; i <= h; i++, k++) v[i] = t[k]; } function mergesort( v, ans, i, j) { if (i < j) { let mid = Math.round((i + j) / 2)-1; // calling mergesort for left half mergesort(v, ans, i, mid); // calling mergesort for right half mergesort(v, ans, mid + 1, j); // merging both halves and generating answer merge(v, ans, i, mid, j); } } function constructLowerArray(arr, n) { let v = []; // inserting elements and corresponding index // as pair for (let i = 0; i < n; i++) { let x = [arr[i], i]; v.push(x); } // answer array for keeping count // initialized by 0, let ans = new Array(n); for(let i = 0; i < n; i++) ans[i] = 0; // calling mergesort mergesort(v, ans, 0, n - 1); return ans; } // Driver Code Starts. let arr= [ 12, 1, 2, 3, 0, 11, 4 ]; let n = arr.length; let ans = constructLowerArray(arr, n); console.log(ans); // This code is contributed by garg28harsh. Time Complexity: O(N * log N), Used for Merge Sort.
Auxiliary Space: O(N), Space used to store the pair values in a different array.
Count smaller elements on the right side using Self-Balancing BST:
The idea is to use a Self Balancing Binary Search Tree (AVL, Red Black,.. etc) can be used to get the solution in O(N log N) time complexity. We can augment these trees so that every node N contains the size of the subtree rooted with N.
Follow the steps below to solve the problem:
- We traverse the array from right to left and insert all elements in an AVL tree.
- While inserting a new key in an AVL tree, first compare the key with the root. If the key is greater than the root, then it is greater than all the nodes in the left subtree of the root.
- So we add the size of the left subtree to the count of smaller elements for the key being inserted.
- We recursively follow the same approach for all nodes down the root.
Below is the implementation of the above approach.
C++ #include <iostream> using namespace std; #include <stdio.h> #include <stdlib.h> // An AVL tree node struct node { int key; struct node* left; struct node* right; int height; // size of the tree rooted // with this node int size; }; // A utility function to get // maximum of two integers int max(int a, int b); // A utility function to get // height of the tree rooted with N int height(struct node* N) { if (N == NULL) return 0; return N->height; } // A utility function to size // of the tree of rooted with N int size(struct node* N) { if (N == NULL) return 0; return N->size; } // A utility function to // get maximum of two integers int max(int a, int b) { return (a > b) ? a : b; } // Helper function that allocates a // new node with the given key and // NULL left and right pointers. struct node* newNode(int key) { struct node* node = (struct node*)malloc(sizeof(struct node)); node->key = key; node->left = NULL; node->right = NULL; // New node is initially added at leaf node->height = 1; node->size = 1; return (node); } // A utility function to right rotate // subtree rooted with y struct node* rightRotate(struct node* y) { struct node* x = y->left; struct node* T2 = x->right; // Perform rotation x->right = y; y->left = T2; // Update heights y->height = max(height(y->left), height(y->right)) + 1; x->height = max(height(x->left), height(x->right)) + 1; // Update sizes y->size = size(y->left) + size(y->right) + 1; x->size = size(x->left) + size(x->right) + 1; // Return new root return x; } // A utility function to left rotate // subtree rooted with x struct node* leftRotate(struct node* x) { struct node* y = x->right; struct node* T2 = y->left; // Perform rotation y->left = x; x->right = T2; // Update heights x->height = max(height(x->left), height(x->right)) + 1; y->height = max(height(y->left), height(y->right)) + 1; // Update sizes x->size = size(x->left) + size(x->right) + 1; y->size = size(y->left) + size(y->right) + 1; // Return new root return y; } // Get Balance factor of node N int getBalance(struct node* N) { if (N == NULL) return 0; return height(N->left) - height(N->right); } // Inserts a new key to the tree rotted with // node. Also, updates *count to contain count // of smaller elements for the new key struct node* insert(struct node* node, int key, int* count) { // 1. Perform the normal BST rotation if (node == NULL) return (newNode(key)); if (key < node->key) node->left = insert(node->left, key, count); else { node->right = insert(node->right, key, count); // UPDATE COUNT OF SMALLER ELEMENTS FOR KEY *count = *count + size(node->left) + 1; } // 2.Update height and size of this ancestor node node->height = max(height(node->left), height(node->right)) + 1; node->size = size(node->left) + size(node->right) + 1; // 3. Get the balance factor of this // ancestor node to check whether this // node became unbalanced int balance = getBalance(node); // If this node becomes unbalanced, // then there are 4 cases // Left Left Case if (balance > 1 && key < node->left->key) return rightRotate(node); // Right Right Case if (balance < -1 && key > node->right->key) return leftRotate(node); // Left Right Case if (balance > 1 && key > node->left->key) { node->left = leftRotate(node->left); return rightRotate(node); } // Right Left Case if (balance < -1 && key < node->right->key) { node->right = rightRotate(node->right); return leftRotate(node); } // Return the (unchanged) node pointer return node; } // The following function updates the // countSmaller array to contain count of // smaller elements on right side. void constructLowerArray(int arr[], int countSmaller[], int n) { int i, j; struct node* root = NULL; // Initialize all the counts in // countSmaller array as 0 for (i = 0; i < n; i++) countSmaller[i] = 0; // Starting from rightmost element, // insert all elements one by one in // an AVL tree and get the count of // smaller elements for (i = n - 1; i >= 0; i--) { root = insert(root, arr[i], &countSmaller[i]); } } // Utility function that prints out an // array on a line void printArray(int arr[], int size) { int i; cout << "\n"; for (i = 0; i < size; i++) cout << arr[i] << " "; } // Driver code int main() { int arr[] = { 12, 1, 2, 3, 0, 11, 4 }; int n = sizeof(arr) / sizeof(arr[0]); int* low = (int*)malloc(sizeof(int) * n); constructLowerArray(arr, low, n); printArray(low, n); return 0; } // This code is contributed by Hemant Jain #include <stdio.h> #include <stdlib.h> // An AVL tree node struct node { int key; struct node* left; struct node* right; int height; int size; // size of the tree rooted with this node }; // A utility function to get maximum of two integers int max(int a, int b); // A utility function to get height of the tree rooted with // N int height(struct node* N) { if (N == NULL) return 0; return N->height; } // A utility function to size of the tree of rooted with N int size(struct node* N) { if (N == NULL) return 0; return N->size; } // A utility function to get maximum of two integers int max(int a, int b) { return (a > b) ? a : b; } /* Helper function that allocates a new node with the given key and NULL left and right pointers. */ struct node* newNode(int key) { struct node* node = (struct node*)malloc(sizeof(struct node)); node->key = key; node->left = NULL; node->right = NULL; node->height = 1; // new node is initially added at leaf node->size = 1; return (node); } // A utility function to right rotate subtree rooted with y struct node* rightRotate(struct node* y) { struct node* x = y->left; struct node* T2 = x->right; // Perform rotation x->right = y; y->left = T2; // Update heights y->height = max(height(y->left), height(y->right)) + 1; x->height = max(height(x->left), height(x->right)) + 1; // Update sizes y->size = size(y->left) + size(y->right) + 1; x->size = size(x->left) + size(x->right) + 1; // Return new root return x; } // A utility function to left rotate subtree rooted with x struct node* leftRotate(struct node* x) { struct node* y = x->right; struct node* T2 = y->left; // Perform rotation y->left = x; x->right = T2; // Update heights x->height = max(height(x->left), height(x->right)) + 1; y->height = max(height(y->left), height(y->right)) + 1; // Update sizes x->size = size(x->left) + size(x->right) + 1; y->size = size(y->left) + size(y->right) + 1; // Return new root return y; } // Get Balance factor of node N int getBalance(struct node* N) { if (N == NULL) return 0; return height(N->left) - height(N->right); } // Inserts a new key to the tree rotted with node. Also, // updates *count to contain count of smaller elements for // the new key struct node* insert(struct node* node, int key, int* count) { /* 1. Perform the normal BST rotation */ if (node == NULL) return (newNode(key)); if (key < node->key) node->left = insert(node->left, key, count); else { node->right = insert(node->right, key, count); // UPDATE COUNT OF SMALLER ELEMENTS FOR KEY *count = *count + size(node->left) + 1; } /* 2. Update height and size of this ancestor node */ node->height = max(height(node->left), height(node->right)) + 1; node->size = size(node->left) + size(node->right) + 1; /* 3. Get the balance factor of this ancestor node to check whether this node became unbalanced */ int balance = getBalance(node); // If this node becomes unbalanced, then there are 4 // cases // Left Left Case if (balance > 1 && key < node->left->key) return rightRotate(node); // Right Right Case if (balance < -1 && key > node->right->key) return leftRotate(node); // Left Right Case if (balance > 1 && key > node->left->key) { node->left = leftRotate(node->left); return rightRotate(node); } // Right Left Case if (balance < -1 && key < node->right->key) { node->right = rightRotate(node->right); return leftRotate(node); } /* return the (unchanged) node pointer */ return node; } // The following function updates the countSmaller array to // contain count of smaller elements on right side. void constructLowerArray(int arr[], int countSmaller[], int n) { int i, j; struct node* root = NULL; // initialize all the counts in countSmaller array as 0 for (i = 0; i < n; i++) countSmaller[i] = 0; // Starting from rightmost element, insert all elements // one by one in an AVL tree and get the count of // smaller elements for (i = n - 1; i >= 0; i--) { root = insert(root, arr[i], &countSmaller[i]); } } /* Utility function that prints out an array on a line */ void printArray(int arr[], int size) { int i; printf("\n"); for (i = 0; i < size; i++) printf("%d ", arr[i]); } // Driver program to test above functions int main() { int arr[] = { 12, 1, 2, 3, 0, 11, 4 }; int n = sizeof(arr) / sizeof(arr[0]); int* low = (int*)malloc(sizeof(int) * n); constructLowerArray(arr, low, n); printArray(low, n); return 0; } import java.util.*; class GFG { // An AVL tree node static class node { int key; node left; node right; int height; // size of the tree rooted // with this node int size; }; static int[] countSmaller; static int count; // A utility function to get // height of the tree rooted with N static int height(node N) { if (N == null) return 0; return N.height; } // A utility function to size // of the tree of rooted with N static int size(node N) { if (N == null) return 0; return N.size; } // A utility function to // get maximum of two integers static int max(int a, int b) { return (a > b) ? a : b; } // Helper function that allocates a // new node with the given key and // null left and right pointers. static node newNode(int key) { node node = new node(); node.key = key; node.left = null; node.right = null; // New node is initially added at leaf node.height = 1; node.size = 1; return (node); } // A utility function to right rotate // subtree rooted with y static node rightRotate(node y) { node x = y.left; node T2 = x.right; // Perform rotation x.right = y; y.left = T2; // Update heights y.height = Math.max(height(y.left), height(y.right)) + 1; x.height = Math.max(height(x.left), height(x.right)) + 1; // Update sizes y.size = size(y.left) + size(y.right) + 1; x.size = size(x.left) + size(x.right) + 1; // Return new root return x; } // A utility function to left rotate // subtree rooted with x static node leftRotate(node x) { node y = x.right; node T2 = y.left; // Perform rotation y.left = x; x.right = T2; // Update heights x.height = Math.max(height(x.left), height(x.right)) + 1; y.height = Math.max(height(y.left), height(y.right)) + 1; // Update sizes x.size = size(x.left) + size(x.right) + 1; y.size = size(y.left) + size(y.right) + 1; // Return new root return y; } // Get Balance factor of node N static int getBalance(node N) { if (N == null) return 0; return height(N.left) - height(N.right); } // Inserts a new key to the tree rotted with // node. Also, updates *count to contain count // of smaller elements for the new key static node insert(node node, int key, int count) { // 1. Perform the normal BST rotation if (node == null) return (newNode(key)); if (key < node.key) node.left = insert(node.left, key, count); else { node.right = insert(node.right, key, count); // UPDATE COUNT OF SMALLER ELEMENTS FOR KEY countSmaller[count] = countSmaller[count] + size(node.left) + 1; } // 2.Update height and size of this ancestor node node.height = Math.max(height(node.left), height(node.right)) + 1; node.size = size(node.left) + size(node.right) + 1; // 3. Get the balance factor of this // ancestor node to check whether this // node became unbalanced int balance = getBalance(node); // If this node becomes unbalanced, // then there are 4 cases // Left Left Case if (balance > 1 && key < node.left.key) return rightRotate(node); // Right Right Case if (balance < -1 && key > node.right.key) return leftRotate(node); // Left Right Case if (balance > 1 && key > node.left.key) { node.left = leftRotate(node.left); return rightRotate(node); } // Right Left Case if (balance < -1 && key < node.right.key) { node.right = rightRotate(node.right); return leftRotate(node); } // Return the (unchanged) node pointer return node; } // The following function updates the // countSmaller array to contain count of // smaller elements on right side. static void constructLowerArray(int arr[], int n) { int i, j; node root = null; // Initialize all the counts in // countSmaller array as 0 for (i = 0; i < n; i++) countSmaller[i] = 0; // Starting from rightmost element, // insert all elements one by one in // an AVL tree and get the count of // smaller elements for (i = n - 1; i >= 0; i--) { root = insert(root, arr[i], i); } } // Utility function that prints out an // array on a line static void printArray(int arr[], int size) { int i; System.out.print("\n"); for (i = 0; i < size; i++) System.out.print(arr[i] + " "); } // Driver code public static void main(String[] args) { int arr[] = { 12, 1, 2, 3, 0, 11, 4 }; int n = arr.length; countSmaller = new int[n]; constructLowerArray(arr, n); printArray(countSmaller, n); } } // This code is contributed by Rajput-Ji import math class GFG: # An AVL tree node class node: key = 0 left = None right = None height = 0 # size of the tree rooted # with this node size = 0 countSmaller = None count = 0 # A utility function to get # height of the tree rooted with N @staticmethod def height(N): if (N == None): return 0 return N.height # A utility function to size # of the tree of rooted with N @staticmethod def size(N): if (N == None): return 0 return N.size # A utility function to # get maximum of two integers @staticmethod def max(a, b): return a if (a > b) else b # Helper function that allocates a # new node with the given key and # null left and right pointers. @staticmethod def newNode(key): node = GFG.node() node.key = key node.left = None node.right = None # New node is initially added at leaf node.height = 1 node.size = 1 return (node) # A utility function to right rotate # subtree rooted with y @staticmethod def rightRotate(y): x = y.left T2 = x.right # Perform rotation x.right = y y.left = T2 # Update heights y.height = max(GFG.height(y.left), GFG.height(y.right)) + 1 x.height = max(GFG.height(x.left), GFG.height(x.right)) + 1 # Update sizes y.size = GFG.size(y.left) + GFG.size(y.right) + 1 x.size = GFG.size(x.left) + GFG.size(x.right) + 1 # Return new root return x # A utility function to left rotate # subtree rooted with x @staticmethod def leftRotate(x): y = x.right T2 = y.left # Perform rotation y.left = x x.right = T2 # Update heights x.height = max(GFG.height(x.left), GFG.height(x.right)) + 1 y.height = max(GFG.height(y.left), GFG.height(y.right)) + 1 # Update sizes x.size = GFG.size(x.left) + GFG.size(x.right) + 1 y.size = GFG.size(y.left) + GFG.size(y.right) + 1 # Return new root return y # Get Balance factor of node N @staticmethod def getBalance(N): if (N == None): return 0 return GFG.height(N.left) - GFG.height(N.right) # Inserts a new key to the tree rotted with # node. Also, updates *count to contain count # of smaller elements for the new key @staticmethod def insert(node, key, count): # 1. Perform the normal BST rotation if (node == None): return (GFG.newNode(key)) if (key < node.key): node.left = GFG.insert(node.left, key, count) else: node.right = GFG.insert(node.right, key, count) # UPDATE COUNT OF SMALLER ELEMENTS FOR KEY GFG.countSmaller[count] = GFG.countSmaller[count] + \ GFG.size(node.left) + 1 # 2.Update height and size of this ancestor node node.height = max(GFG.height(node.left), GFG.height(node.right)) + 1 node.size = GFG.size(node.left) + GFG.size(node.right) + 1 # 3. Get the balance factor of this # ancestor node to check whether this # node became unbalanced balance = GFG.getBalance(node) # If this node becomes unbalanced, # then there are 4 cases # Left Left Case if (balance > 1 and key < node.left.key): return GFG.rightRotate(node) # Right Right Case if (balance < -1 and key > node.right.key): return GFG.leftRotate(node) # Left Right Case if (balance > 1 and key > node.left.key): node.left = GFG.leftRotate(node.left) return GFG.rightRotate(node) # Right Left Case if (balance < -1 and key < node.right.key): node.right = GFG.rightRotate(node.right) return GFG.leftRotate(node) # Return the (unchanged) node pointer return node # The following function updates the # countSmaller array to contain count of # smaller elements on right side. @staticmethod def constructLowerArray(arr, n): i = 0 j = 0 root = None # Initialize all the counts in # countSmaller array as 0 i = 0 while (i < n): GFG.countSmaller[i] = 0 i += 1 # Starting from rightmost element, # insert all elements one by one in # an AVL tree and get the count of # smaller elements i = n - 1 while (i >= 0): root = GFG.insert(root, arr[i], i) i -= 1 # Utility function that prints out an # array on a line @staticmethod def printArray(arr, size): i = 0 print("\n", end="") i = 0 while (i < size): print(str(arr[i]) + " ", end="") i += 1 # Driver code @staticmethod def main(args): arr = [12, 1, 2, 3, 0, 11, 4] n = len(arr) GFG.countSmaller = [0] * (n) GFG.constructLowerArray(arr, n) GFG.printArray(GFG.countSmaller, n) if __name__ == "__main__": GFG.main([]) using System; public class GFG { // An AVL tree node public class node { public int key; public node left; public node right; public int height; // size of the tree rooted // with this node public int size; }; static int[] countSmaller; static int count; // A utility function to get // height of the tree rooted with N static int height(node N) { if (N == null) return 0; return N.height; } // A utility function to size // of the tree of rooted with N static int size(node N) { if (N == null) return 0; return N.size; } // A utility function to // get maximum of two integers static int max(int a, int b) { return (a > b) ? a : b; } // Helper function that allocates a // new node with the given key and // null left and right pointers. static node newNode(int key) { node node = new node(); node.key = key; node.left = null; node.right = null; // New node is initially added at leaf node.height = 1; node.size = 1; return (node); } // A utility function to right rotate // subtree rooted with y static node rightRotate(node y) { node x = y.left; node T2 = x.right; // Perform rotation x.right = y; y.left = T2; // Update heights y.height = Math.Max(height(y.left), height(y.right)) + 1; x.height = Math.Max(height(x.left), height(x.right)) + 1; // Update sizes y.size = size(y.left) + size(y.right) + 1; x.size = size(x.left) + size(x.right) + 1; // Return new root return x; } // A utility function to left rotate // subtree rooted with x static node leftRotate(node x) { node y = x.right; node T2 = y.left; // Perform rotation y.left = x; x.right = T2; // Update heights x.height = Math.Max(height(x.left), height(x.right)) + 1; y.height = Math.Max(height(y.left), height(y.right)) + 1; // Update sizes x.size = size(x.left) + size(x.right) + 1; y.size = size(y.left) + size(y.right) + 1; // Return new root return y; } // Get Balance factor of node N static int getBalance(node N) { if (N == null) return 0; return height(N.left) - height(N.right); } // Inserts a new key to the tree rotted with // node. Also, updates *count to contain count // of smaller elements for the new key static node insert(node node, int key, int count) { // 1. Perform the normal BST rotation if (node == null) return (newNode(key)); if (key < node.key) node.left = insert(node.left, key, count); else { node.right = insert(node.right, key, count); // UPDATE COUNT OF SMALLER ELEMENTS FOR KEY countSmaller[count] = countSmaller[count] + size(node.left) + 1; } // 2.Update height and size of this ancestor node node.height = Math.Max(height(node.left), height(node.right)) + 1; node.size = size(node.left) + size(node.right) + 1; // 3. Get the balance factor of this // ancestor node to check whether this // node became unbalanced int balance = getBalance(node); // If this node becomes unbalanced, // then there are 4 cases // Left Left Case if (balance > 1 && key < node.left.key) return rightRotate(node); // Right Right Case if (balance < -1 && key > node.right.key) return leftRotate(node); // Left Right Case if (balance > 1 && key > node.left.key) { node.left = leftRotate(node.left); return rightRotate(node); } // Right Left Case if (balance < -1 && key < node.right.key) { node.right = rightRotate(node.right); return leftRotate(node); } // Return the (unchanged) node pointer return node; } // The following function updates the // countSmaller array to contain count of // smaller elements on right side. static void constructLowerArray(int[] arr, int n) { int i, j; node root = null; // Initialize all the counts in // countSmaller array as 0 for (i = 0; i < n; i++) countSmaller[i] = 0; // Starting from rightmost element, // insert all elements one by one in // an AVL tree and get the count of // smaller elements for (i = n - 1; i >= 0; i--) { root = insert(root, arr[i], i); } } // Utility function that prints out an // array on a line static void printArray(int[] arr, int size) { int i; Console.Write("\n"); for (i = 0; i < size; i++) Console.Write(arr[i] + " "); } // Driver code public static void Main(String[] args) { int[] arr = { 12, 1, 2, 3, 0, 11, 4 }; int n = arr.Length; countSmaller = new int[n]; constructLowerArray(arr, n); printArray(countSmaller, n); } } // This code is contributed by Rajput-Ji let countSmaller; let count; // An AVL tree node class Node { constructor(key) { this.key = key; this.left = null; this.right = null; this.height = 1; this.size = 1; } } // A utility function to get height of the tree rooted with N function height(N) { if (N === null) return 0; return N.height; } // A utility function to size of the tree of rooted with N function size(N) { if (N === null) return 0; return N.size; } // A utility function to get maximum of two integers function max(a, b) { return a > b ? a : b; } // Helper function that allocates a new node with the given key and null left and right pointers. function newNode(key) { let node = new Node(key); node.key = key; node.left = null; node.right = null; // New node is initially added at leaf node.height = 1; node.size = 1; return node; } // A utility function to right rotate subtree rooted with y function rightRotate(y) { let x = y.left; let T2 = x.right; // Perform rotation x.right = y; y.left = T2; // Update heights y.height = max(height(y.left), height(y.right)) + 1; x.height = max(height(x.left), height(x.right)) + 1; // Update sizes y.size = size(y.left) + size(y.right) + 1; x.size = size(x.left) + size(x.right) + 1; // Return new root return x; } // A utility function to left rotate subtree rooted with x function leftRotate(x) { let y = x.right; let T2 = y.left; // Perform rotation y.left = x; x.right = T2; // Update heights x.height = max(height(x.left), height(x.right)) + 1; y.height = max(height(y.left), height(y.right)) + 1; // Update sizes x.size = size(x.left) + size(x.right) + 1; y.size = size(y.left) + size(y.right) + 1; // Return new root return y; } // Get Balance factor of node N function getBalance(N) { if (N === null) return 0; return height(N.left) - height(N.right); } // Inserts a new key to the tree rotted with node. // Also, updates *count to contain count of smaller elements for the new key function insert(node, key, count) { // 1. Perform the normal BST rotation if (node === null) return newNode(key); if (key < node.key) node.left = insert(node.left, key, count); else { node.right = insert(node.right, key, count); // UPDATE COUNT OF SMALLER ELEMENTS FOR KEY countSmaller[count] = countSmaller[count] + size(node.left) + 1; } // 2.Update height and size of this ancestor node node.height = max(height(node.left), height(node.right)) + 1; node.size = size(node.left) + size(node.right) + 1; // 3. Get the balance factor of this ancestor node to check // whether this node became unbalanced let balance = getBalance(node); // If this node becomes unbalanced, then there are 4 cases // Left Left Case if (balance > 1 && key < node.left.key) { return rightRotate(node); } // Right Right Case if (balance < -1 && key > node.right.key) { return leftRotate(node); } // Left Right Case if (balance > 1 && key > node.left.key) { node.left = leftRotate(node.left); return rightRotate(node); } // Right Left Case if (balance < -1 && key < node.right.key) { node.right = rightRotate(node.right); return leftRotate(node); } // Return the (unchanged) node pointer return node; } // Function to initialize countSmaller array function initializeCountSmaller(arr, n) { countSmaller = new Array(n); // Initialize all element counts as 0 for (let i = 0; i < n; i++) { countSmaller[i] = 0; } // Construct and insert all elements in AVL // tree one by one. Doing this updates // countSmaller[] for each inserted item let root = null; for (let i = n - 1; i >= 0; i--) { root = insert(root, arr[i], i); } } // Function to print countSmaller array function printCountSmaller(arr, n) { for (let i = 0; i < n; i++) { document.write(countSmaller[i] + " "); } } // Example usage let arr = [12, 1, 2, 3, 0, 11, 4]; let n = arr.length; countSmaller = new Array(n); initializeCountSmaller(arr, n); printCountSmaller(arr, n); Time Complexity: O(N * log N), Used self-balancing bst which takes logN time for every operation.
Auxiliary Space: O(N), Space used to store the nodes
Count smaller elements on the right side using BST with 2 extra fields:
The idea is 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.
Follow the steps below to solve the problem:
- Traverse the input array from the ending to the beginning and add the elements into the BST.
- While inserting the elements into the BST, compute the number of elements that are lesser elements simply by computing the sum of the frequency of the element
- And the number of elements to the left side of the current node
- if we are moving to the right side of the current node.
Below is the implementation of the above approach:
C++ #include <bits/stdc++.h> using namespace std; // BST node structure class Node { public: int val; int count; Node* left; Node* right; // Constructor Node(int num1, int num2) { this->val = num1; this->count = num2; this->left = this->right = NULL; } }; // Function to addNode and find the smaller // elements on the right side int addNode(Node*& root, int value, int countSmaller) { // Base case if (root == NULL) { root = new Node(value, 0); return countSmaller; } if (root->val < value) { return root->count + addNode(root->right, value, countSmaller + 1); } else { root->count++; return addNode(root->left, value, countSmaller); } } // Driver code int main() { ios_base::sync_with_stdio(false); cin.tie(0); int data[] = { 12, 1, 2, 3, 0, 11, 4 }; int size = sizeof(data) / sizeof(data[0]); int ans[size] = { 0 }; Node* root = NULL; for (int i = size - 1; i >= 0; i--) { ans[i] = addNode(root, data[i], 0); } for (int i = 0; i < size; i++) cout << ans[i] << " "; return 0; } // This code is contributed by divyanshu gupta class Node { int val; int elecount; int lcount; Node left, right; Node(int val) { this.val = val; this.elecount = 1; this.lcount = 0; this.left = null; this.right = null; } } class Tree { Node root; Tree(Node root) { this.root = root; } int insert(Node node) { int cnt = 0; Node curr = root, prev = null; while (curr != null) { prev = curr; if (node.val > curr.val) { cnt += curr.elecount + curr.lcount; curr = curr.right; } else if (node.val < curr.val) { curr.lcount++; curr = curr.left; } else { prev = curr; prev.elecount++; break; } } if (prev.val > node.val) { prev.left = node; } else if (prev.val < node.val) { prev.right = node; } else { return cnt + prev.lcount - 1; } return cnt; } } public class Main { public static void main(String[] args) { int[] data = { 12, 1, 2, 3, 0, 11, 4 }; int n = data.length; int[] ans = new int[n]; Tree tree = new Tree(new Node(data[n-1])); for (int i = n-2; i >= 0; i--) { ans[i] = tree.insert(new Node(data[i])); } for (int i = 0; i < n; i++) { System.out.print(ans[i] + " "); } } } 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 = [12, 1, 2, 3, 0, 11, 4] print(" ".join(list(map(str, constructArray(arr, n))))) if __name__ == "__main__": main() # Code Contributed by Tarun Gudipati // C# program to implement above logic using System; // BST node structure class GFG { class Node { public int val; public int count; public Node left; public Node right; // Constructor public Node(int num1, int num2) { val = num1; count = num2; left = right = null; } } static Node root = null; // Function to addNode and find the smaller // elements on the right side int addNode(ref Node root, int value, int countSmaller) { // Base case if (root == null) { root = new Node(value, 0); return countSmaller; } if (root.val < value) { return root.count + addNode(ref root.right, value, countSmaller + 1); } else { root.count += 1; return addNode(ref root.left, value, countSmaller); } } // Driver code public static void Main(String[] args) { GFG tree = new GFG(); int[] data = { 12, 1, 2, 3, 0, 11, 4 }; int size = data.Length; int[] ans = new int[size]; for (int i = size - 1; i >= 0; i--) { ans[i] = tree.addNode(ref root, data[i], 0); } for (int i = 0; i < size; i++) Console.Write(ans[i] + " "); } } // This code is contributed by Abhijeet Kumar(abhijeet19403) // BST node structure class Node { constructor(val) { this.val = val; this.left = null; this.right = null; this.elecount = 1; this.lcount = 0; } } class Tree { constructor(root) { this.root = root; } // Function to addNode and find the smaller // elements on the right side insert(node) { let curr = this.root; let cnt = 0; let prev = null; while (curr != null) { prev = curr; if (node.val > curr.val) { cnt += (curr.elecount + curr.lcount); curr = curr.right; } else if (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; } else if (prev.val < node.val) { prev.right = node; } else { return cnt + prev.lcount; } return cnt; } } function constructArray(arr, n) { let t = new Tree(new Node(arr[n - 1])); let ans = [0]; for (let i = n - 2; i >= 0; i--) { ans.push(t.insert(new Node(arr[i]))); } return ans.reverse(); } function main() { let n = 7; let arr = [12, 1, 2, 3, 0, 11, 4]; console.log(constructArray(arr, n).join(" ")); } main(); // This code is contributed by akashish__ Time Complexity: O(N2) as adding step can take O(N) time.
Auxiliary Space: O(N), Space used to store the elements.
Count smaller elements on right side using Set in C++ STL
Here's a simple and concise approach to solve the above problem.
Follow the steps below to solve the problem:
- We start iterating from the end of the array to the beginning.
- For each element, we find the index where it can be inserted in a sorted array of all the elements to its right. This is done using the lower_bound function.
- The number of elements to the right of the current element that are smaller than it is the same as the index where it can be inserted in the sorted array.
- We insert the current element in the sorted array at the appropriate index using the insert function.
- We store the number of smaller elements for the current element in a separate list.
- Finally, we return the list of counts, which is the desired output.
Below is the implementation of above approach
C++ #include <bits/stdc++.h> using namespace std; class Solution { public: vector<int> constructLowerArray(vector<int>& arr) { vector<int> ans, temp; int n = arr.size(); for (int i = n - 1; i >= 0; i--) { int c = lower_bound(temp.begin(), temp.end(), arr[i]) - temp.begin(); ans.push_back(c); temp.insert(temp.begin() + c, arr[i]); } reverse(ans.begin(), ans.end()); return ans; } }; int main() { vector<int> arr = {12, 1, 2, 3, 0, 11, 4}; Solution obj; vector<int> ans = obj.constructLowerArray(arr); for (int x : ans) cout << x << " "; // Output: 6 1 1 1 0 1 0 return 0; } import java.util.*; class Solution { public List<Integer> constructLowerArray(int[] arr) { List<Integer> ans = new ArrayList<>(); List<Integer> temp = new ArrayList<>(); int n = arr.length; for (int i = n - 1; i >= 0; i--) { int c = Collections.binarySearch(temp, arr[i]); if (c < 0) c = -c - 1; ans.add(c); temp.add(c, arr[i]); } Collections.reverse(ans); return ans; } } class Main { public static void main(String[] args) { int[] arr = {12, 1, 2, 3, 0, 11, 4}; Solution obj = new Solution(); List<Integer> ans = obj.constructLowerArray(arr); for (int x : ans) System.out.print(x + " "); // Output: 6 1 1 1 0 1 0 } } import bisect class Solution: def constructLowerArray(self, arr, n): ans = [] temp = [] for i in range(n - 1, -1, -1): c = bisect.bisect_left(temp, arr[i]) ans.append(c) temp.insert(c, arr[i]) return ans[::-1] arr = [12, 1, 2, 3, 0, 11, 4] n = len(arr) s = Solution() ans = s.constructLowerArray(arr, n) print(*ans)
using System; class Solution { public int[] ConstructLowerArray(int[] arr, int n) { int[] ans = new int[n]; int[] temp = new int[n]; for (int i = n - 1; i >= 0; i--) { int c = Array.BinarySearch(temp, 0, n - i - 1, arr[i]); if (c < 0) { c = ~c; } ans[i] = c; Array.Copy(temp, c, temp, c + 1, n - i - c - 1); temp[c] = arr[i]; } return ans; } } class Program { static void Main(string[] args) { int[] arr = {12, 1, 2, 3, 0, 11, 4}; int n = arr.Length; Solution s = new Solution(); int[] ans = s.ConstructLowerArray(arr, n); Console.Write(string.Join(" ", ans)); } } class Solution { constructLowerArray(arr, n) { let ans = [], temp = []; for (let i = n - 1; i >= 0; i--) { let c = temp.findIndex((val) => val >= arr[i]); ans.push(c === -1 ? temp.length : c); temp.splice(c === -1 ? temp.length : c, 0, arr[i]); } return ans.reverse(); } } let obj = new Solution(); let arr = [12, 1, 2, 3, 0, 11, 4]; let n = arr.length; let ans = obj.constructLowerArray(arr, n); console.log(ans); // Output: [6, 1, 1, 1, 0, 1, 0] Time complexity is O(n2)
The space complexity is also O(n) because of the use of two lists ans and temp with n elements.
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