Minimum time to return array to its original state after given modifications

Given two arrays of integers arr and P such that after a cycle an element arr[i] will be at location arr[P[i]]. The task is to find the minimum number of cycles after that all the elements of the array have returned to their original locations.

Examples:

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

Approach: This problem seems to be a typical mathematical problem but if we carefully observe it then we will find that we have to find only the permutation cycles i.e. connected components (formed by the cycles from element movements) and number of nodes in each connected component will represent the time in which the integers will be back in it’s original position for that particular cycle.
For overall graph, do the LCM of the count of nodes from every connected component which is the answer.



Below is the implementation of above approach:

C++

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// 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;
}

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Java

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// 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

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Python

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# 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

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C#

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// 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

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Output:

2


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