Given an array arr[0..n-1] of positive and negative numbers we need to find if there is a cycle in array with given rules of movements. If a number at an i index is positive, then move arr[i]%n forward steps, i.e., next index to visit is (i + arr[i])%n. Conversely, if it’s negative, move backward arr[i]%n steps i.e., next index to visit is (i – arr[i])%n. Here n is size of array. If value of arr[i]%n is zero, then it means no move from index i.

Examples:

Input: arr[] = {2, -1, 1, 2, 2} Output: Yes Explanation: There is a loop in this array because 0 moves to 2, 2 moves to 3, and 3 moves to 0. Input : arr[] = {1, 1, 1, 1, 1, 1} Output : Yes Whole array forms a loop. Input : arr[] = {1, 2} Output : No We move from 0 to index 1. From index 1, there is no move as 2%n is 0. Note that n is 2.

The idea is to form a directed graph of array elements using given set of rules. While forming the graph we don’t make self loops as value arr[i]%n equals to 0 means no moves. Finally our task reduces to detecting cycle in a directed graph. For detecting cycle, we use DFS and in DFS if reach a node which is visited and recursion call stack, we say there is a cycle.

// C++ program to check if a given array is cyclic or not #include<bits/stdc++.h> using namespace std; // A simple Graph DFS based recursive function to check if // there is cycle in graph with vertex v as root of DFS. // Refer below article for details. // http://www.geeksforgeeks.org/detect-cycle-in-a-graph/ bool isCycleRec(int v, vector<int>adj[], vector<bool> &visited, vector<bool> &recur) { visited[v] = true; recur[v] = true; for (int i=0; i<adj[v].size(); i++) { if (visited[adj[v][i]] == false) { if (isCycleRec(adj[v][i], adj, visited, recur)) return true; } // There is a cycle if an adjacent is visited // and present in recursion call stack recur[] else if (visited[adj[v][i]] == true && recur[adj[v][i]] == true) return true; } recur[v] = false; return false; } // Returns true if arr[] has cycle bool isCycle(int arr[], int n) { // Create a graph using given moves in arr[] vector<int>adj[n]; for (int i=0; i<n; i++) adj[i].push_back((i+arr[i]+n)%n); // Do DFS traversal of graph to detect cycle vector<bool> visited(n, false); vector<bool> recur(n, false); for (int i=0; i<n; i++) if (visited[i]==false) if (isCycleRec(i, adj, visited, recur)) return true; return true; } // Driver code int main(void) { int arr[] = {2, -1, 1, 2, 2}; int n = sizeof(arr)/sizeof(arr[0]); if (isCycle(arr, n)) cout << "Yes"<<endl; else cout << "No"<<endl; return 0; }

Output :

Yes

This article is contributed by **Roshni Agarwal**. 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.

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