Check loop in array according to given constraints

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.


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.

Note that self loops are not considered a cycle. For example {0} is not cyclic.

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
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.
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[]
    for (int i=0; i<n; i++)
      if (i != (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;
        cout << "No"<<endl;
    return 0;


Output :


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