# Count smaller elements on right side

Hello Everyone,

Write a function to count number of smaller elements on right of each element in an array. Given an unsorted array arr[] of distinct integers, construct another array countSmaller[] such that countSmaller[i] contains count of smaller elements on right side of each element arr[i] in array.

Examples:

Input: arr[] = {12, 1, 2, 3, 0, 11, 4} Output: countSmaller[] = {6, 1, 1, 1, 0, 1, 0} (Corner Cases) 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}

Method 1 (Simple)
Use two 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[].

`#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, 10, 5, 4, 2,`

` ` `20, 6, 1, 0, 2 };`

` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);`

` ` `int` `*low = (` `int` `*)` `malloc` `(` `sizeof` `(` `int` `)*n);`

` `

` ` `constructLowerArray(arr, low, n);`

` ` `printArray(low, n);`

` `

` ` `return` `0;`

`}`

Output:

8 7 5 4 2 4 3 1 0 0

Time Complexity: O(n^2)
Auxiliary Space: O(1)

Method 2 (Use Self Balancing BST)
A Self Balancing Binary Search Tree (AVL, Red Black,… etc) can be used to get the solution in O(nLogn) time complexity. We can augment these trees so that every node N contains size the subtree rooted with N. We have used AVL tree in the following implementation.
We traverse the array from right to left and insert all elements one by one in an AVL tree. While inserting a new key in an AVL tree, we first compare the key with root. If key is greater than root, then it is greater than all the nodes in left subtree of root. So we add the size of left subtree to the count of smaller element for the key being inserted. We recursively follow the same approach for all nodes down the root.

Following is the C implementation.

• 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[] = {10, 6, 15, 20, 30, 5, 7};`

` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);`

` `

` ` `int` `*low = (` `int` `*)` `malloc` `(` `sizeof` `(` `int` `)*n);`

` `

` ` `constructLowerArray(arr, low, n);`

` `

` ` `cout <<` `"Following is the constructed smaller count array"` `;`

` ` `printArray(low, n);`

` `

` ` `return` `0;`

`}`

Output:

Following is the constructed smaller count array 3 1 2 2 2 0 0

Time Complexity: O(nLogn)
Auxiliary Space: O(n)