Check if the end of the Array can be reached from a given position

/*
https://www.geeksforgeeks.org/check-if-the-end-of-the-array-can-be-reached-from-a-given-position/
This whole code is GFG article code so don't write this code anywhere,Modify the code.

Logic: BFS
We are starting from S and there is two situation only either go s+arr[i] or s-arr[i].so now,we are having two new starting points if any of the above two reaches to end,then answer will be yes.Suppose we again reach at s+arr[i] after some steps then the results will be same.So keep a visited array where we don't need to come to the same point because nothing is going to change,you will follow the same path again you went earlier.
T.C : O(N) as we move all i once.
S.C : O(N) for queue for BFS
*/
def solve(arr, n, start):
    # Simple list used as queue
    q = []

    # Visited array
    visited = [False] * n

    # Push starting index
    q.append(start)

    reached = False

    while len(q) > 0:
        temp = q[0]
        q = q[1:]   # pop front

        # Skip if already visited
        if visited[temp]:
            continue

        visited[temp] = True

        # Check if reached end
        if temp == n - 1:
            reached = True
            break

        # Forward move
        if temp + arr[temp] < n:
            q.append(temp + arr[temp])

        # Backward move
        if temp - arr[temp] >= 0:
            q.append(temp - arr[temp])

    if reached:
        print("Yes")
    else:
        print("No")