Dadas dos permutaciones P1 y P2 de números de 1 a N , la tarea es encontrar el recuento máximo de los mismos elementos correspondientes en las permutaciones dadas realizando un desplazamiento cíclico hacia la izquierda o hacia la derecha en P1 .
Ejemplos:
Entrada: P1 = [5 4 3 2 1], P2 = [1 2 3 4 5]
Salida: 1
Explicación:
Tenemos un par coincidente en el índice 2 para el elemento 3.
Entrada: P1 = [1 3 5 2 4 6] , P2 = [1 5 2 4 3 6]
Salida: 3
Explicación:
el desplazamiento cíclico de la segunda permutación hacia la derecha daría 6 1 5 2 4 3, y obtenemos una coincidencia de 5, 2, 4. Por lo tanto, la respuesta es 3 parejas coincidentes.
Enfoque ingenuo: El enfoque ingenuo consiste en verificar cada cambio posible en la dirección izquierda y derecha, contar el número de pares coincidentes recorriendo todas las permutaciones formadas.
Complejidad de tiempo: O(N 2 )
Espacio auxiliar: O(1)
Enfoque eficiente: El enfoque ingenuo anterior se puede optimizar. La idea es que cada elemento almacene la menor distancia entre las posiciones de este elemento desde los lados izquierdo y derecho en arrays separadas. Por lo tanto, la solución al problema se calculará como la frecuencia máxima de un elemento de las dos arrays separadas. A continuación se muestran los pasos:
- Almacene la posición de todos los elementos de la permutación P2 en una array (digamos store[] ).
- Para cada elemento en la permutación P1 , haga lo siguiente:
- Encuentre la diferencia (digamos diff ) entre la posición del elemento actual en P2 con la posición en P1 .
- Si diff es menor que 0, actualice diff a (N – diff) .
- Almacene la frecuencia de la diferencia actual en un mapa .
- Después de los pasos anteriores, la frecuencia máxima almacenada en el mapa es el número máximo de elementos iguales después de la rotación en P1 .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to maximize the matching
// pairs between two permutation
// using left and right rotation
int maximumMatchingPairs(int perm1[],
int perm2[],
int n)
{
// Left array store distance of element
// from left side and right array store
// distance of element from right side
int left[n], right[n];
// Map to store index of elements
map<int, int> mp1, mp2;
for (int i = 0; i < n; i++) {
mp1[perm1[i]] = i;
}
for (int j = 0; j < n; j++) {
mp2[perm2[j]] = j;
}
for (int i = 0; i < n; i++) {
// idx1 is index of element
// in first permutation
// idx2 is index of element
// in second permutation
int idx2 = mp2[perm1[i]];
int idx1 = i;
if (idx1 == idx2) {
// If element if present on same
// index on both permutations then
// distance is zero
left[i] = 0;
right[i] = 0;
}
else if (idx1 < idx2) {
// Calculate distance from left
// and right side
left[i] = (n - (idx2 - idx1));
right[i] = (idx2 - idx1);
}
else {
// Calculate distance from left
// and right side
left[i] = (idx1 - idx2);
right[i] = (n - (idx1 - idx2));
}
}
// Maps to store frequencies of elements
// present in left and right arrays
map<int, int> freq1, freq2;
for (int i = 0; i < n; i++) {
freq1[left[i]]++;
freq2[right[i]]++;
}
int ans = 0;
for (int i = 0; i < n; i++) {
// Find maximum frequency
ans = max(ans, max(freq1[left[i]],
freq2[right[i]]));
}
// Return the result
return ans;
}
// Driver Code
int main()
{
// Given permutations P1 and P2
int P1[] = { 5, 4, 3, 2, 1 };
int P2[] = { 1, 2, 3, 4, 5 };
int n = sizeof(P1) / sizeof(P1[0]);
// Function Call
cout << maximumMatchingPairs(P1, P2, n);
return 0;
}
Java
// Java program for
// the above approach
import java.util.*;
class GFG{
// Function to maximize the matching
// pairs between two permutation
// using left and right rotation
static int maximumMatchingPairs(int perm1[],
int perm2[],
int n)
{
// Left array store distance of element
// from left side and right array store
// distance of element from right side
int []left = new int[n];
int []right = new int[n];
// Map to store index of elements
HashMap<Integer,
Integer> mp1 = new HashMap<>();
HashMap<Integer,
Integer> mp2 = new HashMap<>();
for (int i = 0; i < n; i++)
{
mp1.put(perm1[i], i);
}
for (int j = 0; j < n; j++)
{
mp2.put(perm2[j], j);
}
for (int i = 0; i < n; i++)
{
// idx1 is index of element
// in first permutation
// idx2 is index of element
// in second permutation
int idx2 = mp2.get(perm1[i]);
int idx1 = i;
if (idx1 == idx2)
{
// If element if present on same
// index on both permutations then
// distance is zero
left[i] = 0;
right[i] = 0;
}
else if (idx1 < idx2)
{
// Calculate distance from left
// and right side
left[i] = (n - (idx2 - idx1));
right[i] = (idx2 - idx1);
}
else
{
// Calculate distance from left
// and right side
left[i] = (idx1 - idx2);
right[i] = (n - (idx1 - idx2));
}
}
// Maps to store frequencies of elements
// present in left and right arrays
HashMap<Integer,
Integer> freq1 = new HashMap<>();
HashMap<Integer,
Integer> freq2 = new HashMap<>();
for (int i = 0; i < n; i++)
{
if(freq1.containsKey(left[i]))
freq1.put(left[i],
freq1.get(left[i]) + 1);
else
freq1.put(left[i], 1);
if(freq2.containsKey(right[i]))
freq2.put(right[i],
freq2.get(right[i]) + 1);
else
freq2.put(right[i], 1);
}
int ans = 0;
for (int i = 0; i < n; i++)
{
// Find maximum frequency
ans = Math.max(ans,
Math.max(freq1.get(left[i]),
freq2.get(right[i])));
}
// Return the result
return ans;
}
// Driver Code
public static void main(String[] args)
{
// Given permutations P1 and P2
int P1[] = {5, 4, 3, 2, 1};
int P2[] = {1, 2, 3, 4, 5};
int n = P1.length;
// Function Call
System.out.print(maximumMatchingPairs(P1, P2, n));
}
}
// This code is contributed by gauravrajput1
Python3
# Python3 program for the above approach
from collections import defaultdict
# Function to maximize the matching
# pairs between two permutation
# using left and right rotation
def maximumMatchingPairs(perm1, perm2, n):
# Left array store distance of element
# from left side and right array store
# distance of element from right side
left = [0] * n
right = [0] * n
# Map to store index of elements
mp1 = {}
mp2 = {}
for i in range (n):
mp1[perm1[i]] = i
for j in range (n):
mp2[perm2[j]] = j
for i in range (n):
# idx1 is index of element
# in first permutation
# idx2 is index of element
# in second permutation
idx2 = mp2[perm1[i]]
idx1 = i
if (idx1 == idx2):
# If element if present on same
# index on both permutations then
# distance is zero
left[i] = 0
right[i] = 0
elif (idx1 < idx2):
# Calculate distance from left
# and right side
left[i] = (n - (idx2 - idx1))
right[i] = (idx2 - idx1)
else :
# Calculate distance from left
# and right side
left[i] = (idx1 - idx2)
right[i] = (n - (idx1 - idx2))
# Maps to store frequencies of elements
# present in left and right arrays
freq1 = defaultdict (int)
freq2 = defaultdict (int)
for i in range (n):
freq1[left[i]] += 1
freq2[right[i]] += 1
ans = 0
for i in range( n):
# Find maximum frequency
ans = max(ans, max(freq1[left[i]],
freq2[right[i]]))
# Return the result
return ans
# Driver Code
if __name__ == "__main__":
# Given permutations P1 and P2
P1 = [ 5, 4, 3, 2, 1 ]
P2 = [ 1, 2, 3, 4, 5 ]
n = len(P1)
# Function Call
print(maximumMatchingPairs(P1, P2, n))
# This code is contributed by chitranayal
C#
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to maximize the matching
// pairs between two permutation
// using left and right rotation
static int maximumMatchingPairs(int []perm1,
int []perm2,
int n)
{
// Left array store distance of element
// from left side and right array store
// distance of element from right side
int []left = new int[n];
int []right = new int[n];
// Map to store index of elements
Dictionary<int,
int> mp1=new Dictionary<int,
int>();
Dictionary<int,
int> mp2=new Dictionary<int,
int>();
for (int i = 0; i < n; i++)
{
mp1[perm1[i]] = i;
}
for (int j = 0; j < n; j++)
{
mp2[perm2[j]] = j;
}
for (int i = 0; i < n; i++)
{
// idx1 is index of element
// in first permutation
// idx2 is index of element
// in second permutation
int idx2 = mp2[perm1[i]];
int idx1 = i;
if (idx1 == idx2)
{
// If element if present on same
// index on both permutations then
// distance is zero
left[i] = 0;
right[i] = 0;
}
else if (idx1 < idx2)
{
// Calculate distance from left
// and right side
left[i] = (n - (idx2 - idx1));
right[i] = (idx2 - idx1);
}
else
{
// Calculate distance from left
// and right side
left[i] = (idx1 - idx2);
right[i] = (n - (idx1 - idx2));
}
}
// Maps to store frequencies of elements
// present in left and right arrays
Dictionary<int,
int> freq1=new Dictionary <int,
int>();
Dictionary<int,
int> freq2=new Dictionary <int,
int>();
for (int i = 0; i < n; i++)
{
if(freq1.ContainsKey(left[i]))
freq1[left[i]]++;
else
freq1[left[i]] = 1;
if(freq2.ContainsKey(right[i]))
freq2[right[i]]++;
else
freq2[right[i]] = 1;
}
int ans = 0;
for (int i = 0; i < n; i++)
{
// Find maximum frequency
ans = Math.Max(ans,
Math.Max(freq1[left[i]],
freq2[right[i]]));
}
// Return the result
return ans;
}
// Driver Code
public static void Main(string[] args)
{
// Given permutations P1 and P2
int []P1 = {5, 4, 3, 2, 1};
int []P2 = {1, 2, 3, 4, 5};
int n = P1.Length;
// Function Call
Console.Write(maximumMatchingPairs(P1, P2, n));
}
}
// This code is contributed by Rutvik_56
Javascript
<script>
// Javascript program for the above approach
// Function to maximize the matching
// pairs between two permutation
// using left and right rotation
function maximumMatchingPairs(perm1, perm2, n)
{
// Left array store distance of element
// from left side and right array store
// distance of element from right side
var left = Array(n);
var right = Array(n);
// Map to store index of elements
var mp1 = new Map(), mp2 = new Map();
for (var i = 0; i < n; i++) {
mp1.set(perm1[i], i);
}
for (var j = 0; j < n; j++) {
mp2.set(perm2[j], j);
}
for (var i = 0; i < n; i++) {
// idx1 is index of element
// in first permutation
// idx2 is index of element
// in second permutation
var idx2 = mp2.get(perm1[i]);
var idx1 = i;
if (idx1 == idx2) {
// If element if present on same
// index on both permutations then
// distance is zero
left[i] = 0;
right[i] = 0;
}
else if (idx1 < idx2) {
// Calculate distance from left
// and right side
left[i] = (n - (idx2 - idx1));
right[i] = (idx2 - idx1);
}
else {
// Calculate distance from left
// and right side
left[i] = (idx1 - idx2);
right[i] = (n - (idx1 - idx2));
}
}
// Maps to store frequencies of elements
// present in left and right arrays
var freq1 = new Map(), freq2 = new Map();
for (var i = 0; i < n; i++) {
if(freq1.has(left[i]))
freq1.set(left[i], freq1.get(left[i])+1)
else
freq1.set(left[i], 1)
if(freq2.has(right[i]))
freq2.set(right[i], freq2.get(right[i])+1)
else
freq2.set(right[i], 1)
}
var ans = 0;
for (var i = 0; i < n; i++) {
// Find maximum frequency
ans = Math.max(ans, Math.max(freq1.get(left[i]),
freq2.get(right[i])));
}
// Return the result
return ans;
}
// Driver Code
// Given permutations P1 and P2
var P1 = [5, 4, 3, 2, 1];
var P2 = [1, 2, 3, 4, 5];
var n = P1.length;
// Function Call
document.write( maximumMatchingPairs(P1, P2, n));
</script>
Producción:
1
Complejidad Temporal: O(NlogN)
Espacio Auxiliar: O(N), ya que se ha tomado N espacio extra.
Publicación traducida automáticamente
Artículo escrito por durgeshkumar30508 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA