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.
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++
// 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. 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++) if (i != (i+arr[i]+n)%n) 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; } |
Python3
# Python3 program to check if a
# given array is cyclic or not
# 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.
# https:#www.geeksforgeeks.org/detect-cycle-in-a-graph/
def isCycleRec(v, adj, visited, recur):
visited[v] = True
recur[v] = True
for i in range(len(adj[v])):
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[]
elif (visited[adj[v][i]] == True and
recur[adj[v][i]] == True):
return True
recur[v] = False
return False
# Returns true if arr[] has cycle
def isCycle(arr, n):
# Create a graph using given
# moves in arr[]
adj = [[] for i in range(n)]
for i in range(n):
if (i != (i + arr[i] + n) % n):
adj[i].append((i + arr[i] + n) % n)
# Do DFS traversal of graph
# to detect cycle
visited = [False] * n
recur = [False] * n
for i in range(n):
if (visited[i] == False):
if (isCycleRec(i, adj,
visited, recur)):
return True
return True
# Driver code
if __name__ == ‘__main__’:
arr = [2, -1, 1, 2, 2]
n = len(arr)
if (isCycle(arr, n)):
print(“Yes”)
else:
print(“No”)
# This code is contributed by PranchalK
Output :
Yes
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