Tiempo mínimo para devolver la array a su estado original después de las modificaciones dadas

Dadas dos arrays de enteros arr y P tales que después de un ciclo un elemento arr[i] estará en la ubicación arr[P[i]] . La tarea es encontrar el número mínimo de ciclos después de que todos los elementos de la array hayan regresado a sus ubicaciones originales.
Ejemplos: 
 

Input: arr[] = {1, 2, 3}, P[] = {3, 2, 1}
Output: 2
After first move array will be {3, 2, 1}
after second move array will be {1, 2, 3}

Input: arr[] = {4, 5, 1, 2, 3}, P[] = {1, 4, 2, 5, 3}
Output: 4

Enfoque: este problema parece ser un problema matemático típico, pero si lo observamos cuidadosamente, encontraremos que solo tenemos que encontrar los ciclos de permutación, es decir, los componentes conectados (formados por los ciclos de los movimientos de los elementos) y el número de Nodes en cada componente conectado. representará el tiempo en que los enteros volverán a su posición original para ese ciclo en particular. 
Para el gráfico general, haga el LCM del conteo de Nodes de cada componente conectado que es la respuesta.
A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return
// lcm of two numbers
int lcm(int a, int b)
{
    return (a * b) / (__gcd(a, b));
}
 
int dfs(int src, vector<int> adj[], vector<bool> &visited)
{
    visited[src] = true;
    int count = 1;
 
    for (int i = 0; i < adj[src].size(); i++)
        if (!visited[adj[src][i]])       
            count +=  dfs(adj[src][i], adj, visited);
   
    return count;
}
 
int findMinTime(int arr[], int P[], int n)
{
    // Make a graph
    vector<int> adj[n+1];
    for (int i = 0; i < n; i++) {
 
        // Add edge
        adj[arr[i]].push_back(P[i]);
    }
 
    // Count reachable nodes from every node.
    vector<bool> visited(n+1);
    int ans = 1;
    for (int i = 0; i < n; i++) {
        if (!visited[i]) {
            ans = lcm(ans, dfs(i, adj, visited));
        }
    }
    return ans;
}
 
// Driver code
int main()
{
 
    int arr[] = { 1, 2, 3 };
    int P[] = { 3, 2, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << findMinTime(arr, P, n);
    return 0;
}

// Java implementation of above approach
import java.util.*;
 
class GFG
{
 
// Function to return
// lcm of two numbers
static int lcm(int a, int b)
{
    return (a * b) / (__gcd(a, b));
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a:__gcd(b, a % b);    
}
 
static int dfs(int src, Vector<Integer> adj[], boolean []visited)
{
    visited[src] = true;
    int count = 1;
 
    for (int i = 0; i < adj[src].size(); i++)
        if (!visited[adj[src].get(i)])    
            count += dfs(adj[src].get(i), adj, visited);
     
    return count;
}
 
static int findMinTime(int arr[], int P[], int n)
{
    // Make a graph
    Vector<Integer> []adj = new Vector[n + 1];
    for (int i = 0; i < n + 1; i++)
        adj[i] = new Vector<Integer>();
    for (int i = 0; i < n; i++)
    {
 
        // Add edge
        adj[arr[i]].add(P[i]);
    }
 
    // Count reachable nodes from every node.
    boolean []visited = new boolean[n + 1];
    int ans = 1;
    for (int i = 0; i < n; i++)
    {
        if (!visited[i])
        {
            ans = lcm(ans, dfs(i, adj, visited));
        }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
 
    int arr[] = { 1, 2, 3 };
    int P[] = { 3, 2, 1 };
    int n = arr.length;
 
    System.out.print(findMinTime(arr, P, n));
}
}
 
// This code is contributed by Rajput-Ji

# Python implementation of above approach
import math
 
# Function to return
# lcm of two numbers
def lcm(a, b):
    return (a * b) // (math.gcd(a, b))
 
def dfs(src, adj, visited):
     
    visited[src] = True
    count = 1
    if adj[src] != 0:
        for i in range(len(adj[src])):
            if (not visited[adj[src][i]]):
                count += dfs(adj[src][i], adj, visited)
     
    return count
 
def findMinTime(arr, P, n):
 
    # Make a graph
    adj = [0] * (n + 1)
    for i in range(n):
 
        # Add edge
        if adj[arr[i]] == 0:
            adj[arr[i]] = []
        adj[arr[i]].append(P[i])
         
    # Count reachable nodes from every node.
    visited = [0] * (n + 1)
    ans = 1
    for i in range(n):
        if (not visited[i]):
            ans = lcm(ans, dfs(i, adj, visited))
     
    return ans
 
# Driver code
 
arr = [1, 2, 3]
P= [3, 2, 1]
n = len(arr)
print(findMinTime(arr, P, n))
 
# This code is contributed by shubhamsingh10

// C# implementation of above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to return
// lcm of two numbers
static int lcm(int a, int b)
{
    return (a * b) / (__gcd(a, b));
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a:__gcd(b, a % b);    
}
 
static int dfs(int src, List<int> []adj, bool []visited)
{
    visited[src] = true;
    int count = 1;
 
    for (int i = 0; i < adj[src].Count; i++)
        if (!visited[adj[src][i]])    
            count += dfs(adj[src][i], adj, visited);
     
    return count;
}
 
static int findMinTime(int []arr, int []P, int n)
{
    // Make a graph
    List<int> []adj = new List<int>[n + 1];
    for (int i = 0; i < n + 1; i++)
        adj[i] = new List<int>();
    for (int i = 0; i < n; i++)
    {
 
        // Add edge
        adj[arr[i]].Add(P[i]);
    }
 
    // Count reachable nodes from every node.
    bool []visited = new bool[n + 1];
    int ans = 1;
    for (int i = 0; i < n; i++)
    {
        if (!visited[i])
        {
            ans = lcm(ans, dfs(i, adj, visited));
        }
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
 
    int []arr = { 1, 2, 3 };
    int []P = { 3, 2, 1 };
    int n = arr.Length;
 
    Console.Write(findMinTime(arr, P, n));
}
}
 
// This code is contributed by 29AjayKumar

<script>
// Javascript implementation of above approach
 
// Function to return
// lcm of two numbers
function lcm(a, b) {
    return (a * b) / (__gcd(a, b));
}
 
function __gcd(a, b) {
    return b == 0 ? a : __gcd(b, a % b);
}
 
function dfs(src, adj, visited) {
    visited[src] = true;
    let count = 1;
 
    for (let i = 0; i < adj[src].length; i++)
        if (!visited[adj[src][i]])
            count += dfs(adj[src][i], adj, visited);
 
    return count;
}
 
function findMinTime(arr, P, n) {
    // Make a graph
    let adj = new Array(n + 1);
    for (let i = 0; i < n + 1; i++)
        adj[i] = new Array();
    for (let i = 0; i < n; i++) {
 
        // Add edge
        adj[arr[i]].push(P[i]);
    }
 
    // Count reachable nodes from every node.
    let visited = new Array(n + 1);
    let ans = 1;
    for (let i = 0; i < n; i++) {
        if (!visited[i]) {
            ans = lcm(ans, dfs(i, adj, visited));
        }
    }
    return ans;
}
 
// Driver code
 
let arr = [1, 2, 3];
let P = [3, 2, 1];
let n = arr.length;
 
document.write(findMinTime(arr, P, n));
 
// This code is contributed by _saurabh_jaiswal
</script>

2