Please select the true options from given list. Thanks.

When storing a stack as an array, the big-o for the push operation uses amortized analysis to get O(n) instead of O(n^2). To get 0(1) for dequeue when using arrays, we need two indexes: one for the front and one for the back. When doubling a circular array queue the array contents do not change their indexes (e.g. the item at index 0 in the smaller array will always end up at index 0 in the larger array). For optimal Big-O when implementing a stack with a singly linked list, the top of the stack is at the "head" of the linked list and the bottom of the stack is at the "tail" of the linked list. For optimal Big-O when implementing a queue with a singly linked list, the front of the queue is at the "head" of the linked list and the back of the queue is at the "tail" of the linked list