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.
#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)