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.

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

GATE CS Corner    Company Wise Coding Practice





Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.