# Count smaller elements on right side

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}
```

## We strongly recommend that you click here and practice it, before moving on to the solution.

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[].

## C

```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, 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;
}
```

## Java

```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, 10, 5, 4, 2, 20, 6, 1, 0, 2};
int n = arr.length;
int low[] = new int[n];
small.constructLowerArray(arr, low, n);
small.printArray(low, n);
}
}

```

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 C implementation.

```#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[] = {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);

printf("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)

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3.8 Average Difficulty : 3.8/5.0
Based on 111 vote(s)