Dadas dos arrays A[] y B[] que constan de N enteros positivos y una array List[][] que consta de M pares de índices, la tarea es minimizar el recuento de elementos desiguales con el mismo índice ( A i != B i ) de las dos arrays intercambiando entre cualquier par de índices dados en la array A[].
Ejemplos:
Entrada: N = 5, M = 4, A[] = {1, 5, 9, 2, 3}, B[] = {2, 4, 5, 1, 3}, Lista[][] = {{ 1, 4}, {2, 3}, {3, 5}, {2, 5}}
Salida: 1
Explicación:
array inicial A[] = {1, 5, 9, 2, 3}
Índices de intercambio (1, 4) en A[] = {2, 5, 9, 1, 3}
Intercambio de índices (2, 3) en A[] = {2, 9, 5, 1, 3}
Por lo tanto, al comparar el arreglo final A[ ] (= {2, 9, 5, 1, 3}) con la array B[] (= {2, 4, 5, 1, 3}), el único par de elementos desiguales con el mismo índice es A[2] ! = B[2] (indexación basada en 1)Entrada: N = 5, M = 4, A[] = {1, 10, 19, 6, 2}, B[] = {1, 6, 10, 2, 19}, Lista[][] = {{ 1, 4}, {2, 3}, {3, 5}, {2, 5}}
Salida: 2
Enfoque: dado que los elementos de la array A[] especificados en la lista de pares se pueden intercambiar cualquier número de veces, esos elementos se pueden considerar como un conjunto conectado con intercambios dentro de sus elementos permitidos. A continuación se detallan los pasos para implementar este enfoque:
- Recorra la array dada A[] y cree un conjunto de componentes conectados que se pueden hacer fácilmente usando el conjunto disjunto .
- Para cada elemento en B[] encuentre si el elemento correspondiente en A[] (A[i], B[i]) pertenece o no al mismo componente conexo.
- En caso afirmativo, haga A[i]=B[i] y continúe. De lo contrario, incremente el recuento de pares no coincidentes.
- Imprima el recuento de pares no coincidentes después de todos los pasos anteriores.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
int find(int par[],int x);
void unionn(int par[], int a, int b);
// Function that count of the mismatched
// pairs in bot the array
int countPairs(int A[], int B[], int N,
int M, int List[][2])
{
int count = 0;
// Create a parent array
// and initialize it
int par[N + 1];
for(int i = 0; i <= N; i++)
par[i] = i;
// Preprocessing of the given
// pairs of indices
for(int i = 0; i < M; i++)
{
// 1-based indexing
int index1 = find(par, List[i][0] - 1);
int index2 = find(par, List[i][1] - 1);
// If both indices doesn't belong
// to same component
if (index1 != index2)
{
// Insert the indices in
// same component
unionn(par, index1, index2);
}
}
// Map to get the indices
// of array A
map<int, int> mp;
for(int i = 0; i < N; i++)
{
mp[A[i]] = i;
}
for(int i = 0; i < N; i++)
{
if (A[i] != B[i])
{
// If current element is not
// present in array B then
// count this as mismatched
if(mp.find(B[i]) == mp.end())
{
count++;
continue;
}
// Get the index of the element
// in array B
int j = mp[B[i]];
// Check if both indices belong
// to same connected component
// if not increment the count
if (find(par, i) != find(par, j))
count++;
}
}
// Return answer
return count;
}
// Function that gives the connected
// component of each index
int find(int par[],int x)
{
if (par[x] == x)
return x;
else
return par[x] = find(par, par[x]);
}
// Function that creates the
// connected components
void unionn(int par[], int a, int b)
{
// Find parent of a and b
// recursively
a = find(par, a);
b = find(par, b);
if (a == b)
return;
// Update the parent of a
par[a] = b;
}
// Driver Code
int main()
{
int N = 5;
int M = 4;
// Given arrays A[], B[]
int A[] = { 1, 5, 9, 2, 3 };
int B[] = { 2, 4, 5, 1, 3 };
// List of indices
int List[][2] = { { 1, 4 }, { 2, 3 },
{ 3, 5 }, { 2, 5 } };
// Function call
cout << countPairs(A, B, N, M, List);
return 0;
}
// This code is contributed by rutvik_56
Java
// Java program for the above approach
import java.util.*;
class GFG {
// Function that count of the mismatched
// pairs in bot the array
public static int countPairs(int A[], int B[],
int N, int M,
int[][] List)
{
int count = 0;
// Create a parent array
// and initialize it
int par[] = new int[N + 1];
for (int i = 0; i <= N; i++)
par[i] = i;
// Preprocessing of the given
// pairs of indices
for (int i = 0; i < M; i++) {
// 1-based indexing
int index1
= find(par, List[i][0] - 1);
int index2
= find(par, List[i][1] - 1);
// If both indices doesn't belong
// to same component
if (index1 != index2) {
// Insert the indices in
// same component
union(par, index1, index2);
}
}
// HashMap to get the indices
// of array A
HashMap<Integer, Integer> map
= new HashMap<>();
for (int i = 0; i < N; i++) {
map.put(A[i], i);
}
for (int i = 0; i < N; i++) {
if (A[i] != B[i]) {
// If current element is not
// present in array B then
// count this as mismatched
if (!map.containsKey(B[i])) {
count++;
continue;
}
// Get the index of the element
// in array B
int j = map.get(B[i]);
// Check if both indices belong
// to same connected component
// if not increment the count
if (find(par, i) != find(par, j))
count++;
}
}
// Return answer
return count;
}
// Function that gives the connected
// component of each index
public static int find(int par[],
int x)
{
if (par[x] == x)
return x;
else
return par[x]
= find(par, par[x]);
}
// Function that creates the
// connected components
public static void
union(int par[], int a, int b)
{
// Find parent of a and b
// recursively
a = find(par, a);
b = find(par, b);
if (a == b)
return;
// Update the parent of a
par[a] = b;
}
// Driver Code
public static void
main(String[] args)
{
int N = 5;
int M = 4;
// Given arrays A[], B[]
int A[] = { 1, 5, 9, 2, 3 };
int B[] = { 2, 4, 5, 1, 3 };
// List of indices
int List[][]
= { { 1, 4 }, { 2, 3 },
{ 3, 5 }, { 2, 5 } };
// Function Call
System.out.println(
countPairs(A, B, N, M, List));
}
}
Python3
# Python3 program for the above approach
# Function that count of the mismatched
# pairs in bot the array
def countPairs(A, B, N, M, List):
count = 0
# Create a parent array
# and initialize it
par = [0] * (N + 1)
for i in range(N + 1):
par[i] = i
# Preprocessing of the given
# pairs of indices
for i in range(M):
# 1-based indexing
index1 = find(par, List[i][0] - 1)
index2 = find(par, List[i][1] - 1)
# If both indices doesn't belong
# to same component
if(index1 != index2):
# Insert the indices in
# same component
union(par, index1, index2)
# HashMap to get the indices
# of array A
map = {}
for i in range(N):
map[A[i]] = i
for i in range(N):
if(A[i] != B[i]):
# If current element is not
# present in array B then
# count this as mismatched
if(B[i] not in map.keys()):
count += 1
continue
# Get the index of the element
# in array B
j = map[B[i]]
# Check if both indices belong
# to same connected component
# if not increment the count
if(find(par, i) != find(par, j)):
count += 1
# Return answer
return count
# Function that gives the connected
# component of each index
def find(par, x):
if(par[x] == x):
return x
else:
par[x] = find(par, par[x])
return par[x]
# Function that creates the
# connected components
def union(par, a, b):
# Find parent of a and b
# recursively
a = find(par, a)
b = find(par, b)
if(a == b):
return
# Update the parent of a
par[a] = b
# Driver Code
N = 5
M = 4
# Given arrays A[], B[]
A = [ 1, 5, 9, 2, 3 ]
B = [ 2, 4, 5, 1, 3 ]
# List of indices
List = [ [ 1, 4 ], [ 2, 3 ],
[ 3, 5 ], [ 2, 5 ] ]
# Function call
print(countPairs(A, B, N, M, List))
# This code is contributed by Shivam Singh
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function that count of the mismatched
// pairs in bot the array
public static int countPairs(int[] A, int[] B, int N,
int M, int[, ] List)
{
int count = 0;
// Create a parent array
// and initialize it
int[] par = new int[N + 1];
for (int i = 0; i <= N; i++)
par[i] = i;
// Preprocessing of the given
// pairs of indices
for (int i = 0; i < M; i++)
{
// 1-based indexing
int index1 = find(par, List[i, 0] - 1);
int index2 = find(par, List[i, 1] - 1);
// If both indices doesn't belong
// to same component
if (index1 != index2)
{
// Insert the indices in
// same component
union(par, index1, index2);
}
}
// Dictionary to get the indices
// of array A
Dictionary<int, int> map = new Dictionary<int, int>();
for (int i = 0; i < N; i++)
{
if (map.ContainsKey(A[i]))
map[A[i]] = i;
else
map.Add(A[i], i);
}
for (int i = 0; i < N; i++)
{
if (A[i] != B[i])
{
// If current element is not
// present in array B then
// count this as mismatched
if (!map.ContainsKey(B[i]))
{
count++;
continue;
}
// Get the index of the element
// in array B
int j = map[B[i]];
// Check if both indices belong
// to same connected component
// if not increment the count
if (find(par, i) != find(par, j))
count++;
}
}
// Return answer
return count;
}
// Function that gives the connected
// component of each index
public static int find(int[] par, int x)
{
if (par[x] == x)
return x;
else
return par[x] = find(par, par[x]);
}
// Function that creates the
// connected components
public static void union(int[] par, int a, int b)
{
// Find parent of a and b
// recursively
a = find(par, a);
b = find(par, b);
if (a == b)
return;
// Update the parent of a
par[a] = b;
}
// Driver Code
public static void Main(String[] args)
{
int N = 5;
int M = 4;
// Given arrays []A, []B
int[] A = {1, 5, 9, 2, 3};
int[] B = {2, 4, 5, 1, 3};
// List of indices
int[, ] List = {{1, 4}, {2, 3}, {3, 5}, {2, 5}};
// Function Call
Console.WriteLine(countPairs(A, B, N, M, List));
}
}
// This code is contributed by shikhasingrajput
Javascript
<script>
// Javascript program for the above approach
// Function that count of the mismatched
// pairs in bot the array
function countPairs(A, B, N, M, List)
{
var count = 0;
// Create a parent array
// and initialize it
var par = Array(N+1);
for(var i = 0; i <= N; i++)
par[i] = i;
// Preprocessing of the given
// pairs of indices
for(var i = 0; i < M; i++)
{
// 1-based indexing
var index1 = find(par, List[i][0] - 1);
var index2 = find(par, List[i][1] - 1);
// If both indices doesn't belong
// to same component
if (index1 != index2)
{
// Insert the indices in
// same component
unionn(par, index1, index2);
}
}
// Map to get the indices
// of array A
var mp = new Map();
for(var i = 0; i < N; i++)
{
mp.set(A[i], i);
}
for(var i = 0; i < N; i++)
{
if (A[i] != B[i])
{
// If current element is not
// present in array B then
// count this as mismatched
if(!mp.has(B[i]))
{
count++;
continue;
}
// Get the index of the element
// in array B
var j = mp.get(B[i]);
// Check if both indices belong
// to same connected component
// if not increment the count
if (find(par, i) != find(par, j))
count++;
}
}
// Return answer
return count;
}
// Function that gives the connected
// component of each index
function find(par,x)
{
if (par[x] == x)
return x;
else
return par[x] = find(par, par[x]);
}
// Function that creates the
// connected components
function unionn(par, a, b)
{
// Find parent of a and b
// recursively
a = find(par, a);
b = find(par, b);
if (a == b)
return;
// Update the parent of a
par[a] = b;
}
// Driver Code
var N = 5;
var M = 4;
// Given arrays A[], B[]
var A = [1, 5, 9, 2, 3];
var B = [2, 4, 5, 1, 3];
// List of indices
var List = [ [ 1, 4 ], [ 2, 3 ],
[ 3, 5 ], [ 2, 5 ] ];
// Function call
document.write( countPairs(A, B, N, M, List));
// This code is contributed by itsok.
</script>
Producción:
1
Complejidad temporal: O(N + M)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por hemanthswarna1506 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA