El conteo de inversión para una array indica qué tan lejos (o cerca) está la array de ser ordenada. Si una array ya está ordenada, el recuento de inversión es 0. Si una array se ordena en orden inverso, el recuento de inversión es el máximo.
Dos elementos a[i] y a[j] forman una inversión si a[i] > a[j] e i < j. Para simplificar, podemos suponer que todos los elementos son únicos.
Ejemplo:
Input: arr[] = {8, 4, 2, 1}
Output: 6
Explanation: Given array has six inversions:
(8,4), (4,2),(8,2), (8,1), (4,1), (2,1).
Input: arr[] = {3, 1, 2}
Output: 2
Explanation:Given array has two inversions:
(3, 1), (3, 2)
Ya hemos discutido el enfoque Naive y los enfoques basados en Merge Sort para contar inversiones .
Análisis de complejidad de la solución en la publicación mencionada anteriormente:
- La complejidad temporal del enfoque Naive es O(n 2 )
- La complejidad temporal del enfoque basado en la ordenación por fusión es O(n Log n).
Revise el árbol AVL antes de leer este artículo.
Hay un enfoque más eficiente para resolver el problema.
Enfoque: la idea es utilizar el árbol de búsqueda binaria autoequilibrado como el árbol rojo-negro, el árbol AVL , etc. y aumentarlo para que cada Node también realice un seguimiento del número de Nodes en el subárbol derecho. Entonces, cada Node contendrá el recuento de Nodes en su subárbol derecho, es decir, el número de Nodes mayor que ese número. Entonces, se puede ver que el conteo aumenta cuando hay un par (a,b) , donde a aparece antes que b en la array y a > b . Entonces, a medida que la array se recorre desde el principio hasta el final, agregue los elementos a la El árbol AVL y el conteo de los Nodes en su subárbol derecho del Node recién insertado será el conteo aumentado o el número de pares (a,b) donde b es el elemento presente.
Algoritmo:
- Cree un árbol AVL , con la propiedad de que cada Node contendrá el tamaño de su subárbol.
- Atraviesa la array de principio a fin.
- Para cada elemento, inserte el elemento en el árbol AVL
- El recuento de los Nodes que son mayores que el elemento actual se puede averiguar comprobando el tamaño del subárbol de sus hijos derechos, por lo que se puede garantizar que los elementos en el subárbol derecho del Node actual tienen un índice menor que el elemento actual y sus valores son mayores que el elemento actual. Entonces esos elementos satisfacen los criterios.
- Por lo tanto, aumente el recuento por tamaño del subárbol del hijo derecho del Node insertado actual.
- Mostrar el conteo.
Implementación:
C++
// An AVL Tree based C++ program to count
// inversion in an array
#include<bits/stdc++.h>
using namespace std;
// An AVL tree node
struct Node
{
int key, height;
struct Node *left, *right;
// size of the tree rooted with this Node
int size;
};
// A utility function to get the 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;
}
/* 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 = new Node;
node->key = key;
node->left = node->right = NULL;
node->height = 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
// *result (inversion count)
struct Node* insert(struct Node* node, int key, int *result)
{
/* 1. Perform the normal BST rotation */
if (node == NULL)
return(newNode(key));
if (key < node->key)
{
node->left = insert(node->left, key, result);
// UPDATE COUNT OF GREATE ELEMENTS FOR KEY
*result = *result + size(node->right) + 1;
}
else
node->right = insert(node->right, key, result);
/* 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 returns inversion count in arr[]
int getInvCount(int arr[], int n)
{
struct Node *root = NULL; // Create empty AVL Tree
int result = 0; // Initialize result
// Starting from first element, insert all elements one by
// one in an AVL tree.
for (int i=0; i<n; i++)
// Note that address of result is passed as insert
// operation updates result by adding count of elements
// greater than arr[i] on left of arr[i]
root = insert(root, arr[i], &result);
return result;
}
// Driver program to test above
int main()
{
int arr[] = {8, 4, 2, 1};
int n = sizeof(arr)/sizeof(int);
cout << "Number of inversions count are : "
<< getInvCount(arr,n);
return 0;
}
Java
// AVL Tree based Java program to count
// inversion in an array
import java.util.*;
class GfG{
// Initialize result
static int result = 0;
// An AVL tree node
static class Node
{
int key, height;
Node left, right;
// Size of the tree rooted
// with this Node
int size;
}
// A utility function to get the 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 create a new node
static Node newNode(int ele)
{
Node temp = new Node();
temp.key = ele;
temp.left = null;
temp.right = null;
temp.height = 1;
temp.size = 1;
return temp;
}
// 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 *result
// (inversion count)
static Node insert(Node node, int key)
{
// 1. Perform the normal BST rotation
if (node == null)
return (newNode(key));
if (key < node.key)
{
node.left = insert(node.left, key);
// UPDATE COUNT OF GREATER ELEMENTS FOR KEY
result = result + size(node.right) + 1;
}
else
node.right = insert(node.right, key);
// 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 returns inversion
// count in arr[]
static void getInvCount(int arr[], int n)
{
// Create empty AVL Tree
Node root = null;
// Starting from first element,
// insert all elements one by
// one in an AVL tree.
for(int i = 0; i < n; i++)
// Note that address of result
// is passed as insert operation
// updates result by adding count
// of elements greater than arr[i]
// on left of arr[i]
root = insert(root, arr[i]);
}
// Driver code
public static void main(String[] args)
{
int[] arr = new int[] { 8, 4, 2, 1 };
int n = arr.length;
getInvCount(arr, n);
System.out.print("Number of inversions " +
"count are : " + result);
}
}
// This code is contributed by tushar_bansal
Python3
# An AVL Tree based Python program to
# count inversion in an array
# A utility function to get height of
# the tree rooted with N
def height(N):
if N == None:
return 0
return N.height
# A utility function to size of the
# tree of rooted with N
def size(N):
if N == None:
return 0
return N.size
# Helper function that allocates a new
# Node with the given key and NULL left
# and right pointers.
class newNode:
def __init__(self, key):
self.key = key
self.left = self.right = None
self.height = self.size = 1
# A utility function to right rotate
# subtree rooted with y
def rightRotate(y):
x = y.left
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
def leftRotate(x):
y = x.right
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
def getBalance(N):
if N == None:
return 0
return height(N.left) - height(N.right)
# Inserts a new key to the tree rotted
# with Node. Also, updates *result (inversion count)
def insert(node, key, result):
# 1. Perform the normal BST rotation
if node == None:
return newNode(key)
if key < node.key:
node.left = insert(node.left, key, result)
# UPDATE COUNT OF GREATER ELEMENTS FOR KEY
result[0] = result[0] + size(node.right) + 1
else:
node.right = insert(node.right, key, result)
# 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
balance = getBalance(node)
# If this node becomes unbalanced,
# then there are 4 cases
# Left Left Case
if (balance > 1 and key < node.left.key):
return rightRotate(node)
# Right Right Case
if (balance < -1 and key > node.right.key):
return leftRotate(node)
# Left Right Case
if balance > 1 and key > node.left.key:
node.left = leftRotate(node.left)
return rightRotate(node)
# Right Left Case
if balance < -1 and key < node.right.key:
node.right = rightRotate(node.right)
return leftRotate(node)
# return the (unchanged) node pointer
return node
# The following function returns
# inversion count in arr[]
def getInvCount(arr, n):
root = None # Create empty AVL Tree
result = [0] # Initialize result
# Starting from first element, insert all
# elements one by one in an AVL tree.
for i in range(n):
# Note that address of result is passed
# as insert operation updates result by
# adding count of elements greater than
# arr[i] on left of arr[i]
root = insert(root, arr[i], result)
return result[0]
# Driver Code
if __name__ == '__main__':
arr = [8, 4, 2, 1]
n = len(arr)
print("Number of inversions count are :",
getInvCount(arr, n))
# This code is contributed by PranchalK
C#
// AVL Tree based C# program to count
// inversion in an array
using System;
class GfG
{
// Initialize result
static int result = 0;
// An AVL tree node
public
class Node
{
public
int key, height;
public
Node left, right;
// Size of the tree rooted
// with this Node
public
int size;
}
// A utility function to get the 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 create a new node
static Node newNode(int ele)
{
Node temp = new Node();
temp.key = ele;
temp.left = null;
temp.right = null;
temp.height = 1;
temp.size = 1;
return temp;
}
// 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 *result
// (inversion count)
static Node insert(Node node, int key)
{
// 1. Perform the normal BST rotation
if (node == null)
return (newNode(key));
if (key < node.key)
{
node.left = insert(node.left, key);
// UPDATE COUNT OF GREATER ELEMENTS FOR KEY
result = result + size(node.right) + 1;
}
else
node.right = insert(node.right, key);
// 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 returns inversion
// count in []arr
static void getInvCount(int []arr, int n)
{
// Create empty AVL Tree
Node root = null;
// Starting from first element,
// insert all elements one by
// one in an AVL tree.
for(int i = 0; i < n; i++)
// Note that address of result
// is passed as insert operation
// updates result by adding count
// of elements greater than arr[i]
// on left of arr[i]
root = insert(root, arr[i]);
}
// Driver code
public static void Main(String[] args)
{
int[] arr = new int[] { 8, 4, 2, 1 };
int n = arr.Length;
getInvCount(arr, n);
Console.Write("Number of inversions " +
"count are : " + result);
}
}
// This code is contributed by gauravrajput1
Javascript
<script>
// AVL Tree based javascript program to count
// inversion in an
// Initialize result
var result = 0;
// An AVL tree node
class Node {
constructor(){
this.key = 0, this.height = 0;
this.left = null, this.right = null;
// Size of the tree rooted
// with this Node
this.size = 0;
}
}
// A utility function to get the 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 create a new node
function newNode(ele) {
var temp = new Node();
temp.key = ele;
temp.left = null;
temp.right = null;
temp.height = 1;
temp.size = 1;
return temp;
}
// A utility function to right rotate
// subtree rooted with y
function rightRotate(y) {
var x = y.left;
var 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
function leftRotate(x) {
var y = x.right;
var 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
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 *result
// (inversion count)
function insert(node , key) {
// 1. Perform the normal BST rotation
if (node == null)
return (newNode(key));
if (key < node.key) {
node.left = insert(node.left, key);
// UPDATE COUNT OF GREATER ELEMENTS FOR KEY
result = result + size(node.right) + 1;
} else
node.right = insert(node.right, key);
// 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
var 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 returns inversion
// count in arr
function getInvCount(arr , n) {
// Create empty AVL Tree
var root = null;
// Starting from first element,
// insert all elements one by
// one in an AVL tree.
for (i = 0; i < n; i++)
// Note that address of result
// is passed as insert operation
// updates result by adding count
// of elements greater than arr[i]
// on left of arr[i]
root = insert(root, arr[i]);
}
// Driver code
var arr = [ 8, 4, 2, 1 ];
var n = arr.length;
getInvCount(arr, n);
document.write("Number of inversions " +
"count are : " + result);
// This code contributed by aashish1995
</script>
Producción
Number of inversions count are : 6
Análisis de Complejidad:
- Complejidad Temporal: O(n Log n).
La inserción en una inserción AVL toma un tiempo O (log n) y se insertan n elementos en el árbol, por lo que la complejidad del tiempo es O (n log n). - Complejidad espacial: O(n).
Para crear un árbol AVL con un máximo de n Nodes O(n), se requiere espacio adicional.
Contando inversiones usando Set en C++ STL.
Pronto discutiremos el enfoque basado en el árbol indexado binario para el mismo.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA