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:

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Output:

2


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.