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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ implementation of above approach #include 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 adj[], vector &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 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 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);        cout << findMinTime(arr, P, n);     return 0; }

Output:

2

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