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

• Difficulty Level : Medium
• Last Updated : 09 Jun, 2021

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 solve it on “PRACTICE” 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++

 `// 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<``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

 `// 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 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 []adj = ``new` `Vector[n + ``1``];``    ``for` `(``int` `i = ``0``; i < n + ``1``; i++)``        ``adj[i] = ``new` `Vector();``    ``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

 `# 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#

 `// 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`

Javascript

 ``
Output:
`2`

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