El conteo de inversión para una array indica qué tan lejos (o cerca) está la array de ser ordenada. Si la array ya está ordenada, el recuento de inversión es 0. Si la array está ordenada 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.
Entonces, nuestra tarea es contar el número de inversiones en la array. Ese es el número de pares de elementos (a[i], a[j]) tales que:
- a[i] > a[j] y,
- yo < j.
Ejemplo :
Input: arr[] = {8, 4, 2, 1}
Output: 6
Given array has six inversions (8, 4), (4, 2),
(8, 2), (8, 1), (4, 1), (2, 1).
Input: arr[] = { 1, 20, 6, 4, 5 }
Output: 5
Enfoque:
iteraremos hacia atrás en la array y almacenaremos cada elemento en el Trie. Para almacenar un número en Trie,
tenemos que dividir el número en su forma binaria y si el bit es 0 , significa que almacenamos ese bit en el puntero izquierdo del Node actual y si es 1 lo almacenaremos en el puntero derecho del Node actual y cambiar correspondientemente el Node actual. También mantendremos el conteo que significa cuántos números siguen el mismo camino hasta ese Node.
Estructura del Node del Trie
struct node{
int count;
node* left;
node* right;
};
En cualquier momento, mientras almacenamos los bits, nos movemos hacia el puntero derecho (es decir, el bit es 1) verificaremos si el hijo izquierdo existe, entonces esto significa que hay números que son más pequeños que el número actual que ya son sido almacenado en el Trie, estos números son solo el conteo de inversión, por lo que los agregaremos al conteo.
A continuación se muestra la implementación del enfoque.
C++
// C++ implementation
#include <iostream>
using namespace std;
// Structure of the node
struct Node {
int count;
Node* left;
Node* right;
};
// function to initialize
// new node
Node* makeNewNode()
{
Node* temp = new Node;
temp->count = 1;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// Insert element in trie
void insertElement(int num,
Node* root,
int& ans)
{
// Converting the number
// into binary form
for (int i = 63; i >= 0; i--) {
// Checking if the i-th
// bit ios set or not
int a = (num & (1 << i));
// If the bit is 1
if (a != 0) {
// if the bit is 1 that means
// we have to go to the right
// but we also checks if left
// pointer exists i.e there is
// at least a number smaller than
// the current number already in
// the trie we add that count
// to ans
if (root->left != NULL)
ans += root->left->count;
// If right pointer is not NULL
// we just iterate to that
// position and increment the count
if (root->right != NULL) {
root = root->right;
root->count += 1;
}
// If right is NULL we add a new
// node over there and initialize
// the count with 1
else {
Node* temp = makeNewNode();
root->right = temp;
root = root->right;
}
}
// if the bit is 0
else {
// We have to iterate to left,
// we first check if left
// exists? if yes then change
// the root and the count
if (root->left != NULL) {
root = root->left;
root->count++;
}
// otherwise we create
// the left node
else {
Node* temp = makeNewNode();
root->left = temp;
root = root->left;
}
}
}
}
// function to count
// the inversions
int getInvCount(int arr[], int n)
{
Node* head = makeNewNode();
int ans = 0;
for (int i = n - 1; i >= 0; i--) {
// inserting each element in Trie
insertElement(arr[i],
head,
ans);
}
return ans;
}
// Driver Code
int main()
{
int arr[] = { 8, 4, 2, 1 };
int n = sizeof(arr) / sizeof(int);
cout << "Number of inversions are : "
<< getInvCount(arr, n);
return 0;
}
Java
// Java implementation of above idea
import java.util.*;
class GFG
{
// Structure of the node
static class Node
{
int count;
Node left;
Node right;
};
static int ans;
// function to initialize
// new node
static Node makeNewNode()
{
Node temp = new Node();
temp.count = 1;
temp.left = null;
temp.right = null;
return temp;
}
// Insert element in trie
static void insertElement(int num,
Node root)
{
// Converting the number
// into binary form
for (int i = 63; i >= 0; i--)
{
// Checking if the i-th
// bit ios set or not
int a = (num & (1 << i));
// If the bit is 1
if (a != 0)
{
// if the bit is 1 that means
// we have to go to the right
// but we also checks if left
// pointer exists i.e there is
// at least a number smaller than
// the current number already in
// the trie we add that count
// to ans
if (root.left != null)
ans += root.left.count;
// If right pointer is not null
// we just iterate to that
// position and increment the count
if (root.right != null)
{
root = root.right;
root.count += 1;
}
// If right is null we add a new
// node over there and initialize
// the count with 1
else
{
Node temp = makeNewNode();
root.right = temp;
root = root.right;
}
}
// if the bit is 0
else
{
// We have to iterate to left,
// we first check if left
// exists? if yes then change
// the root and the count
if (root.left != null)
{
root = root.left;
root.count++;
}
// otherwise we create
// the left node
else
{
Node temp = makeNewNode();
root.left = temp;
root = root.left;
}
}
}
}
// function to count
// the inversions
static int getInvCount(int arr[], int n)
{
Node head = makeNewNode();
ans = 0;
for (int i = n - 1; i >= 0; i--)
{
// inserting each element in Trie
insertElement(arr[i],
head);
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 8, 4, 2, 1 };
int n = arr.length;
System.out.print("Number of inversions are : "
+ getInvCount(arr, n));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation
# Structure of the node
class Node:
def __init__(self):
self.left = self.right = None
self.count = 1
# function to initialize
# new node
def makeNewNode():
temp = Node()
return temp
# Insert element in trie
def insertElement(num, root, ans):
# Converting the number
# into binary form
for i in range(63, -1, -1):
# Checking if the i-th
# bit ios set or not
a = (num & (1 << i));
# If the bit is 1
if (a != 0):
# if the bit is 1 that means
# we have to go to the right
# but we also checks if left
# pointer exists i.e there is
# at least a number smaller than
# the current number already in
# the trie we add that count
# to ans
if (root.left != None):
ans += root.left.count;
# If right pointer is not None
# we just iterate to that
# position and increment the count
if (root.right != None):
root = root.right;
root.count += 1;
# If right is None we add a new
# node over there and initialize
# the count with 1
else:
temp = makeNewNode();
root.right = temp;
root = root.right;
# if the bit is 0
else:
# We have to iterate to left,
# we first check if left
# exists? if yes then change
# the root and the count
if (root.left != None):
root = root.left;
root.count += 1
# otherwise we create
# the left node
else:
temp = makeNewNode();
root.left = temp;
root = root.left;
return ans
# function to count
# the inversions
def getInvCount(arr, n):
head = makeNewNode();
ans = 0;
for i in range(n - 1 ,-1, -1):
# inserting each element in Trie
ans = insertElement(arr[i], head, ans);
return ans;
# Driver Code
if __name__=='__main__':
arr = [ 8, 4, 2, 1 ]
n = len(arr)
print("Number of inversions are : " + str(getInvCount(arr, n)))
# This code is contributed by rutvik_56
C#
// C# implementation of above idea
using System;
class GFG
{
// Structure of the node
public class Node
{
public int count;
public Node left;
public Node right;
};
static int ans;
// function to initialize
// new node
static Node makeNewNode()
{
Node temp = new Node();
temp.count = 1;
temp.left = null;
temp.right = null;
return temp;
}
// Insert element in trie
static void insertElement(int num,
Node root)
{
// Converting the number
// into binary form
for (int i = 63; i >= 0; i--)
{
// Checking if the i-th
// bit ios set or not
int a = (num & (1 << i));
// If the bit is 1
if (a != 0)
{
// if the bit is 1 that means
// we have to go to the right
// but we also checks if left
// pointer exists i.e there is
// at least a number smaller than
// the current number already in
// the trie we add that count
// to ans
if (root.left != null)
ans += root.left.count;
// If right pointer is not null
// we just iterate to that
// position and increment the count
if (root.right != null)
{
root = root.right;
root.count += 1;
}
// If right is null we add a new
// node over there and initialize
// the count with 1
else
{
Node temp = makeNewNode();
root.right = temp;
root = root.right;
}
}
// if the bit is 0
else
{
// We have to iterate to left,
// we first check if left
// exists? if yes then change
// the root and the count
if (root.left != null)
{
root = root.left;
root.count++;
}
// otherwise we create
// the left node
else
{
Node temp = makeNewNode();
root.left = temp;
root = root.left;
}
}
}
}
// function to count the inversions
static int getInvCount(int []arr, int n)
{
Node head = makeNewNode();
ans = 0;
for (int i = n - 1; i >= 0; i--)
{
// inserting each element in Trie
insertElement(arr[i], head);
}
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 8, 4, 2, 1 };
int n = arr.Length;
Console.Write("Number of inversions are : " +
getInvCount(arr, n));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript implementation of above idea
// Structure of the node
class Node
{
constructor()
{
this.count = 0;
this.left = null;
this.right = null;
}
};
var ans = 0;
// function to initialize
// new node
function makeNewNode()
{
var temp = new Node();
temp.count = 1;
temp.left = null;
temp.right = null;
return temp;
}
// Insert element in trie
function insertElement(num, root)
{
// Converting the number
// into binary form
for (var i = 63; i >= 0; i--)
{
// Checking if the i-th
// bit ios set or not
var a = (num & (1 << i));
// If the bit is 1
if (a != 0)
{
// if the bit is 1 that means
// we have to go to the right
// but we also checks if left
// pointer exists i.e there is
// at least a number smaller than
// the current number already in
// the trie we add that count
// to ans
if (root.left != null)
ans += root.left.count;
// If right pointer is not null
// we just iterate to that
// position and increment the count
if (root.right != null)
{
root = root.right;
root.count += 1;
}
// If right is null we add a new
// node over there and initialize
// the count with 1
else
{
var temp = makeNewNode();
root.right = temp;
root = root.right;
}
}
// if the bit is 0
else
{
// We have to iterate to left,
// we first check if left
// exists? if yes then change
// the root and the count
if (root.left != null)
{
root = root.left;
root.count++;
}
// otherwise we create
// the left node
else
{
var temp = makeNewNode();
root.left = temp;
root = root.left;
}
}
}
}
// function to count the inversions
function getInvCount(arr, n)
{
var head = makeNewNode();
ans = 0;
for (var i = n - 1; i >= 0; i--)
{
// inserting each element in Trie
insertElement(arr[i], head);
}
return ans;
}
// Driver Code
var arr = [8, 4, 2, 1];
var n = arr.length;
document.write("Number of inversions are : " +
getInvCount(arr, n));
</script>
Producción:
Number of inversions are : 6
Tiempo Complejidad: 
Espacio Auxiliar :