<reponame>UPstartDeveloper/CS-1.3-Core-Data-Structures
#!python


class Node(object):

    def __init__(self, data):
        """Initialize this node with the given data."""
        self.data = data
        self.next = None

    def __repr__(self):
        """Return a string representation of this node."""
        return 'Node({!r})'.format(self.data)


class LinkedList(object):

    def __init__(self, iterable=None):
        """Initialize this linked list and append the given items, if any."""
        self.head = None  # First node
        self.tail = None  # Last node
        self.size = 0  # Number of nodes
        # Append the given items
        if iterable is not None:
            for item in iterable:
                self.append(item)

    def __str__(self):
        """Return a formatted string representation of this linked list."""
        items = ['({!r})'.format(item) for item in self.items()]
        return '[{}]'.format(' -> '.join(items))

    def __repr__(self):
        """Return a string representation of this linked list."""
        return 'LinkedList({!r})'.format(self.items())

    def items(self):
        """Return a list of all items in this linked list.
        Best and worst case running time: Theta(n) for n items in the list
        because we always need to loop through all n nodes.

        """
        # Create an empty list of results
        result = []  # Constant time to create a new list
        # Start at the head node
        node = self.head  # Constant time to assign a variable reference
        # Loop until the node is None, which is one node too far past the tail
        while node is not None:  # Always n iterations because no early exit
            # Append this node's data to the results list
            result.append(node.data)  # Constant time to append to a list
            # Skip to the next node
            node = node.next  # Constant time to reassign a variable
        # Now result contains the data from all nodes
        return result  # Constant time to return a list

    def is_empty(self):
        """Return True if this linked list is empty, or False."""
        return self.head is None

    def length(self):
        """Return the length of this linked list by traversing its nodes.
           This method runs in constant time, otherwise represented as O(1), in
           all scenarios. This is because all we do is retrieve the value
           stored in the size property of the given LinkedList instance.

        """
        return self.size

    def get_at_index(self, index):
        """Return the item at the given index in this linked list, or
           raise ValueError if the given index is out of range of the list
           size.

           Best case running time: O(1)
           This method runs in constant time if we are getting the item at the
           index 0 (also can be thought of as the head node), or if the index
           is out of range of the list. In the former, this is because we only
           need one iteration through the for loop. In the latter, it is
           because we can be sure we will need no iteration of the for loop,
           and just raise the ValuError exception.

           Worst case running time: O(n)
           This occurs if we are trying to get the item at the last index
           position in the list, otherwise referred to as the tail node. This
           is because we will need to traverse through all the items in the
           list, which increases in linear time with respect to the growth of
           items in the list, which is represented by n.

        """
        # Check if the given index is out of range and if so raise an error
        if not (0 <= index < self.size):
            raise ValueError('List index out of range: {}'.format(index))
        # Find the node at the given index
        node = self.head
        for i in range(index):
            node = node.next
        # return its data
        return node.data

    def insert_at_index(self, index, item):
        """Insert the given item at the given index in this linked list, or
           raise ValueError if the given index is out of range of the list
           size.

           Best case running time: O(1)
           This function mimics the approach of the prepend and append methods
           in several scenarios: if we are inserting an item at the head node
           position (which is index 0), this uses the runtime of the prepend
           method; or if we inserting at the tail node, or we are inserting an
           item into a list with no other items, this method uses the same
           runtime as the append operation (in the latter case, the index
           would also be equal to 0). Furthermore, this method also runs in
           constant time if the index is out of range, in which it takes
           constant time to raise tbe ValueError.

           Worst case running time: O(n)
           If we are inserting an item at an index position that cannot be
           considered the head nor the tail, then we have a variable number
           of items to iterate over. In the worst case scenario, the index we
           want to insert at is just before the tail, meaning we need n-1
           iterations through the for loop (found in the else block below),
           where n is the number of items in the list. The runtime in this
           scenario scales in linear time to the growth of the list, therefore
           we disregard the -1 and are left with O(n) runtime.

        """
        # Check if the given index is out of range and if so raise an error
        if not (0 <= index <= self.size):
            raise ValueError('List index out of range: {}'.format(index))
        # if we are inserting at the tail, or no previous nodes, then append
        if self.size == index:
            self.append(item)
        # if we are inserting at the head, then just perform a prepend
        elif index == 0:
            self.prepend(item)
        else:
            # Perform traversal of the nodes
            prev_node = self.head
            # find the node at index, as well as the one just before it
            for i in range(index - 1):
                prev_node = prev_node.next
            node_being_moved = prev_node.next
            # make a new Node for the data to be inserted
            new_node = Node(item)
            # insert the new node at the index, and connect it to the rest
            prev_node.next = new_node
            new_node.next = node_being_moved
            # increment the size property
            self.size += 1

    def append(self, item):
        """Insert the given item at the tail of this linked list.

           Best and worst case running time: O(1)
           In all scenarios, this method runs in constant time. The runtime
           complexity of all its statements are independent of the size of the
           list. This is because in all the statements we always instantiate a
           new Node, check the size of the list, rearrange the pointers of the
           head and tail properties (and in some cases the next property of
           the tail). Finally, we end by incrementing the value of the size
           property.

        """
        # Create a new node to hold the given item
        new_node = Node(item)
        # Check if this linked list is empty
        if self.is_empty():
            # Assign head to new node
            self.head = new_node
        else:
            # Otherwise insert new node after tail
            self.tail.next = new_node
        # Update tail to new node regardless
        self.tail = new_node
        # Increment the size property
        self.size += 1

    def prepend(self, item):
        """Insert the given item at the head of this linked list.

           Best and worst case running time: O(1)
           Similar to the append method, the runtime of the prepend operation
           is independent of the size of the list. In all scenarios, we merely
           need to make a new node, rearrange the pointers of the head and
           tail, and increment the size of the list.

        """
        # Create a new node to hold the given item
        new_node = Node(item)
        # Check if this linked list is empty
        if self.is_empty():
            # Assign tail to new node
            self.tail = new_node
        else:
            # Otherwise insert new node before head
            new_node.next = self.head
        # Update head to new node regardless
        self.head = new_node
        # Increment the size property
        self.size += 1

    def find(self, quality):
        """Return an item from this linked list satisfying the given quality.

           Best case running time: O(1) if item is near the head of the list.

           Worst case running time: O(n) if item is near the tail of the list
           or not present and we need to loop through all n nodes in the list.

        """
        # Start at the head node
        node = self.head  # Constant time to assign a variable reference
        # Loop until the node is None, which is one node too far past the tail
        while node is not None:  # Up to n iterations if we don't exit early
            # Check if this node's data satisfies the given quality function
            if quality(node.data):  # Constant time to call quality function
                # We found data satisfying the quality function, so exit early
                return node.data  # Constant time to return data
            # Skip to the next node
            node = node.next  # Constant time to reassign a variable
        # We never found data satisfying quality, but have to return something
        return None  # Constant time to return None

    def replace(self, old_item, new_item):
        """Replace the given old_item in this linked list with given new_item
           using the same node, or raise ValueError if old_item is not found.

           Best case running time: O(1)
           If we are trying to replace an item in a list that's empty, that
           has only one item, or the item is present at the head, we have a
           constant runtime that is independent of how the size of the list.
           In the first scenario mentioned, we would not need any traversal of
           the list; and in the other two scenarios, we would need exactly one.

           Worst case running time: O(n)
           If we are trying to replace an item that is present at the end of
           the list, or not present in the list at all, we will need to
           traverse through all the items in the list. This operation scales in
           linear time, because it grows in direct proportion to the number of
           items in the list (since one iteration is required for each item).

        """
        # Start at the head node
        node = self.head
        # traverse the nodes to find old_item
        while node is not None:
            # Check if this node's data matches the old_item
            if node.data == old_item:
                # We found the node, now change the data it references
                node.data = new_item
                # end the method by returning None
                return None
            else:
                # Skip to the next node
                node = node.next
        # old_item is not in the list, so raise a ValuError
        raise ValueError(f'Item not in list: {old_item}')

    def delete(self, item):
        """Delete the given item from this linked list, or raise ValueError.

           Best case running time: O(1)
           If the item is at the head of the list, then we only require one
           iteration for traversal, and then we rely on decision strucutres to
           appropiately remove the item. The runtime of these operations will
           not change based on the length of the list.

           Worst case running time: O(n)
           If the item at the end of the list, or not present at all, this
           method will have to traverse through all the items in the list to
           get to it. This causes the runtime of this method to reach O(n)
           asymptotically, because it is essentially an implementation of the
           linear search algorithm.

        """
        # Start at the head node
        node = self.head
        # Keep track of the node before the one containing the given item
        previous = None
        # Create a flag to track if we have found the given item
        found = False
        # Loop until we have found the given item or the node is None
        while found is False and node is not None:
            # Check if the node's data matches the given item
            if node.data == item:
                # We found data matching the given item, so update found flag
                found = True
            else:
                # Skip to the next node
                previous = node
                node = node.next
        # Check if we found the given item or we never did and reached the tail
        if found is True:
            # Check if we found a node in the middle of this linked list
            if node is not self.head and node is not self.tail:
                # Update the previous node to skip around the found node
                previous.next = node.next
                # Unlink the found node from its next node
                node.next = None
            # Check if we found a node at the head
            if node is self.head:
                # Update head to the next node
                self.head = node.next
                # Unlink the found node from the next node
                node.next = None
            # Check if we found a node at the tail
            if node is self.tail:
                # Check if there is a node before the found node
                if previous is not None:
                    # Unlink the previous node from the found node
                    previous.next = None
                # Update tail to the previous node regardless
                self.tail = previous
            # Decrement the size property
            self.size -= 1
        else:
            # Otherwise raise an error to tell the user that delete has failed
            raise ValueError('Item not found: {}'.format(item))

    def __iter__(self):
        """Return a generator of the data referenced by the Nodes in this
           LinkedList.

           Credit to my old LinkedList class in my CS 1.2 repo:
           https://github.com/UPstartDeveloper/CS-1.2-Intro-Data-Structures/blob/master/Code/linkedlist.py

        """
        node = self.head
        while node is not None:
            item = node.data
            node = node.next
            yield(item)


def test_linked_list():
    ll = LinkedList()
    print(ll)

    print('Appending items:')
    ll.append('A')
    print(ll)
    ll.append('B')
    print(ll)
    ll.append('C')
    print(ll)
    print('head: {}'.format(ll.head))
    print('tail: {}'.format(ll.tail))
    print('size: {}'.format(ll.size))
    print('length: {}'.format(ll.length()))
    print([item for item in ll])

    print('Getting items by index:')
    for index in range(ll.size):
        item = ll.get_at_index(index)
        print('get_at_index({}): {!r}'.format(index, item))

    print('Deleting items:')
    ll.delete('B')
    print(ll)
    ll.delete('C')
    print(ll)
    ll.delete('A')
    print(ll)
    print('head: {}'.format(ll.head))
    print('tail: {}'.format(ll.tail))
    print('size: {}'.format(ll.size))
    print('length: {}'.format(ll.length()))


if __name__ == '__main__':
    test_linked_list()

#!python


class Node(object):

    def __init__(self, data):
        """Initialize this node with the given data."""
        self.data = data
        self.next = None

    def __repr__(self):
        """Return a string representation of this node."""
        return 'Node({!r})'.format(self.data)


class LinkedList(object):

    def __init__(self, iterable=None):
        """Initialize this linked list and append the given items, if any."""
        self.head = None  # First node
        self.tail = None  # Last node
        self.size = 0  # Number of nodes
        # Append the given items
        if iterable is not None:
            for item in iterable:
                self.append(item)

    def __str__(self):
        """Return a formatted string representation of this linked list."""
        items = ['({!r})'.format(item) for item in self.items()]
        return '[{}]'.format(' -> '.join(items))

    def __repr__(self):
        """Return a string representation of this linked list."""
        return 'LinkedList({!r})'.format(self.items())

    def items(self):
        """Return a list of all items in this linked list.
        Best and worst case running time: Theta(n) for n items in the list
        because we always need to loop through all n nodes.

        """
        # Create an empty list of results
        result = []  # Constant time to create a new list
        # Start at the head node
        node = self.head  # Constant time to assign a variable reference
        # Loop until the node is None, which is one node too far past the tail
        while node is not None:  # Always n iterations because no early exit
            # Append this node's data to the results list
            result.append(node.data)  # Constant time to append to a list
            # Skip to the next node
            node = node.next  # Constant time to reassign a variable
        # Now result contains the data from all nodes
        return result  # Constant time to return a list

    def is_empty(self):
        """Return True if this linked list is empty, or False."""
        return self.head is None

    def length(self):
        """Return the length of this linked list by traversing its nodes.
           This method runs in constant time, otherwise represented as O(1), in
           all scenarios. This is because all we do is retrieve the value
           stored in the size property of the given LinkedList instance.

        """
        return self.size

    def get_at_index(self, index):
        """Return the item at the given index in this linked list, or
           raise ValueError if the given index is out of range of the list
           size.

           Best case running time: O(1)
           This method runs in constant time if we are getting the item at the
           index 0 (also can be thought of as the head node), or if the index
           is out of range of the list. In the former, this is because we only
           need one iteration through the for loop. In the latter, it is
           because we can be sure we will need no iteration of the for loop,
           and just raise the ValuError exception.

           Worst case running time: O(n)
           This occurs if we are trying to get the item at the last index
           position in the list, otherwise referred to as the tail node. This
           is because we will need to traverse through all the items in the
           list, which increases in linear time with respect to the growth of
           items in the list, which is represented by n.

        """
        # Check if the given index is out of range and if so raise an error
        if not (0 <= index < self.size):
            raise ValueError('List index out of range: {}'.format(index))
        # Find the node at the given index
        node = self.head
        for i in range(index):
            node = node.next
        # return its data
        return node.data

    def insert_at_index(self, index, item):
        """Insert the given item at the given index in this linked list, or
           raise ValueError if the given index is out of range of the list
           size.

           Best case running time: O(1)
           This function mimics the approach of the prepend and append methods
           in several scenarios: if we are inserting an item at the head node
           position (which is index 0), this uses the runtime of the prepend
           method; or if we inserting at the tail node, or we are inserting an
           item into a list with no other items, this method uses the same
           runtime as the append operation (in the latter case, the index
           would also be equal to 0). Furthermore, this method also runs in
           constant time if the index is out of range, in which it takes
           constant time to raise tbe ValueError.

           Worst case running time: O(n)
           If we are inserting an item at an index position that cannot be
           considered the head nor the tail, then we have a variable number
           of items to iterate over. In the worst case scenario, the index we
           want to insert at is just before the tail, meaning we need n-1
           iterations through the for loop (found in the else block below),
           where n is the number of items in the list. The runtime in this
           scenario scales in linear time to the growth of the list, therefore
           we disregard the -1 and are left with O(n) runtime.

        """
        # Check if the given index is out of range and if so raise an error
        if not (0 <= index <= self.size):
            raise ValueError('List index out of range: {}'.format(index))
        # if we are inserting at the tail, or no previous nodes, then append
        if self.size == index:
            self.append(item)
        # if we are inserting at the head, then just perform a prepend
        elif index == 0:
            self.prepend(item)
        else:
            # Perform traversal of the nodes
            prev_node = self.head
            # find the node at index, as well as the one just before it
            for i in range(index - 1):
                prev_node = prev_node.next
            node_being_moved = prev_node.next
            # make a new Node for the data to be inserted
            new_node = Node(item)
            # insert the new node at the index, and connect it to the rest
            prev_node.next = new_node
            new_node.next = node_being_moved
            # increment the size property
            self.size += 1

    def append(self, item):
        """Insert the given item at the tail of this linked list.

           Best and worst case running time: O(1)
           In all scenarios, this method runs in constant time. The runtime
           complexity of all its statements are independent of the size of the
           list. This is because in all the statements we always instantiate a
           new Node, check the size of the list, rearrange the pointers of the
           head and tail properties (and in some cases the next property of
           the tail). Finally, we end by incrementing the value of the size
           property.

        """
        # Create a new node to hold the given item
        new_node = Node(item)
        # Check if this linked list is empty
        if self.is_empty():
            # Assign head to new node
            self.head = new_node
        else:
            # Otherwise insert new node after tail
            self.tail.next = new_node
        # Update tail to new node regardless
        self.tail = new_node
        # Increment the size property
        self.size += 1

    def prepend(self, item):
        """Insert the given item at the head of this linked list.

           Best and worst case running time: O(1)
           Similar to the append method, the runtime of the prepend operation
           is independent of the size of the list. In all scenarios, we merely
           need to make a new node, rearrange the pointers of the head and
           tail, and increment the size of the list.

        """
        # Create a new node to hold the given item
        new_node = Node(item)
        # Check if this linked list is empty
        if self.is_empty():
            # Assign tail to new node
            self.tail = new_node
        else:
            # Otherwise insert new node before head
            new_node.next = self.head
        # Update head to new node regardless
        self.head = new_node
        # Increment the size property
        self.size += 1

    def find(self, quality):
        """Return an item from this linked list satisfying the given quality.

           Best case running time: O(1) if item is near the head of the list.

           Worst case running time: O(n) if item is near the tail of the list
           or not present and we need to loop through all n nodes in the list.

        """
        # Start at the head node
        node = self.head  # Constant time to assign a variable reference
        # Loop until the node is None, which is one node too far past the tail
        while node is not None:  # Up to n iterations if we don't exit early
            # Check if this node's data satisfies the given quality function
            if quality(node.data):  # Constant time to call quality function
                # We found data satisfying the quality function, so exit early
                return node.data  # Constant time to return data
            # Skip to the next node
            node = node.next  # Constant time to reassign a variable
        # We never found data satisfying quality, but have to return something
        return None  # Constant time to return None

    def replace(self, old_item, new_item):
        """Replace the given old_item in this linked list with given new_item
           using the same node, or raise ValueError if old_item is not found.

           Best case running time: O(1)
           If we are trying to replace an item in a list that's empty, that
           has only one item, or the item is present at the head, we have a
           constant runtime that is independent of how the size of the list.
           In the first scenario mentioned, we would not need any traversal of
           the list; and in the other two scenarios, we would need exactly one.

           Worst case running time: O(n)
           If we are trying to replace an item that is present at the end of
           the list, or not present in the list at all, we will need to
           traverse through all the items in the list. This operation scales in
           linear time, because it grows in direct proportion to the number of
           items in the list (since one iteration is required for each item).

        """
        # Start at the head node
        node = self.head
        # traverse the nodes to find old_item
        while node is not None:
            # Check if this node's data matches the old_item
            if node.data == old_item:
                # We found the node, now change the data it references
                node.data = new_item
                # end the method by returning None
                return None
            else:
                # Skip to the next node
                node = node.next
        # old_item is not in the list, so raise a ValuError
        raise ValueError(f'Item not in list: {old_item}')

    def delete(self, item):
        """Delete the given item from this linked list, or raise ValueError.

           Best case running time: O(1)
           If the item is at the head of the list, then we only require one
           iteration for traversal, and then we rely on decision strucutres to
           appropiately remove the item. The runtime of these operations will
           not change based on the length of the list.

           Worst case running time: O(n)
           If the item at the end of the list, or not present at all, this
           method will have to traverse through all the items in the list to
           get to it. This causes the runtime of this method to reach O(n)
           asymptotically, because it is essentially an implementation of the
           linear search algorithm.

        """
        # Start at the head node
        node = self.head
        # Keep track of the node before the one containing the given item
        previous = None
        # Create a flag to track if we have found the given item
        found = False
        # Loop until we have found the given item or the node is None
        while found is False and node is not None:
            # Check if the node's data matches the given item
            if node.data == item:
                # We found data matching the given item, so update found flag
                found = True
            else:
                # Skip to the next node
                previous = node
                node = node.next
        # Check if we found the given item or we never did and reached the tail
        if found is True:
            # Check if we found a node in the middle of this linked list
            if node is not self.head and node is not self.tail:
                # Update the previous node to skip around the found node
                previous.next = node.next
                # Unlink the found node from its next node
                node.next = None
            # Check if we found a node at the head
            if node is self.head:
                # Update head to the next node
                self.head = node.next
                # Unlink the found node from the next node
                node.next = None
            # Check if we found a node at the tail
            if node is self.tail:
                # Check if there is a node before the found node
                if previous is not None:
                    # Unlink the previous node from the found node
                    previous.next = None
                # Update tail to the previous node regardless
                self.tail = previous
            # Decrement the size property
            self.size -= 1
        else:
            # Otherwise raise an error to tell the user that delete has failed
            raise ValueError('Item not found: {}'.format(item))

    def __iter__(self):
        """Return a generator of the data referenced by the Nodes in this
           LinkedList.

           Credit to my old LinkedList class in my CS 1.2 repo:
           https://github.com/UPstartDeveloper/CS-1.2-Intro-Data-Structures/blob/master/Code/linkedlist.py

        """
        node = self.head
        while node is not None:
            item = node.data
            node = node.next
            yield(item)


def test_linked_list():
    ll = LinkedList()
    print(ll)

    print('Appending items:')
    ll.append('A')
    print(ll)
    ll.append('B')
    print(ll)
    ll.append('C')
    print(ll)
    print('head: {}'.format(ll.head))
    print('tail: {}'.format(ll.tail))
    print('size: {}'.format(ll.size))
    print('length: {}'.format(ll.length()))
    print([item for item in ll])

    print('Getting items by index:')
    for index in range(ll.size):
        item = ll.get_at_index(index)
        print('get_at_index({}): {!r}'.format(index, item))

    print('Deleting items:')
    ll.delete('B')
    print(ll)
    ll.delete('C')
    print(ll)
    ll.delete('A')
    print(ll)
    print('head: {}'.format(ll.head))
    print('tail: {}'.format(ll.tail))
    print('size: {}'.format(ll.size))
    print('length: {}'.format(ll.length()))


if __name__ == '__main__':
    test_linked_list()
#!python Initialize this node with the given data. Return a string representation of this node. Initialize this linked list and append the given items, if any. # First node # Last node # Number of nodes # Append the given items Return a formatted string representation of this linked list. Return a string representation of this linked list. Return a list of all items in this linked list.         Best and worst case running time: Theta(n) for n items in the list         because we always need to loop through all n nodes. # Create an empty list of results # Constant time to create a new list # Start at the head node # Constant time to assign a variable reference # Loop until the node is None, which is one node too far past the tail # Always n iterations because no early exit # Append this node's data to the results list # Constant time to append to a list # Skip to the next node # Constant time to reassign a variable # Now result contains the data from all nodes # Constant time to return a list Return True if this linked list is empty, or False. Return the length of this linked list by traversing its nodes.            This method runs in constant time, otherwise represented as O(1), in            all scenarios. This is because all we do is retrieve the value            stored in the size property of the given LinkedList instance. Return the item at the given index in this linked list, or            raise ValueError if the given index is out of range of the list            size.             Best case running time: O(1)            This method runs in constant time if we are getting the item at the            index 0 (also can be thought of as the head node), or if the index            is out of range of the list. In the former, this is because we only            need one iteration through the for loop. In the latter, it is            because we can be sure we will need no iteration of the for loop,            and just raise the ValuError exception.             Worst case running time: O(n)            This occurs if we are trying to get the item at the last index            position in the list, otherwise referred to as the tail node. This            is because we will need to traverse through all the items in the            list, which increases in linear time with respect to the growth of            items in the list, which is represented by n. # Check if the given index is out of range and if so raise an error # Find the node at the given index # return its data Insert the given item at the given index in this linked list, or            raise ValueError if the given index is out of range of the list            size.             Best case running time: O(1)            This function mimics the approach of the prepend and append methods            in several scenarios: if we are inserting an item at the head node            position (which is index 0), this uses the runtime of the prepend            method; or if we inserting at the tail node, or we are inserting an            item into a list with no other items, this method uses the same            runtime as the append operation (in the latter case, the index            would also be equal to 0). Furthermore, this method also runs in            constant time if the index is out of range, in which it takes            constant time to raise tbe ValueError.             Worst case running time: O(n)            If we are inserting an item at an index position that cannot be            considered the head nor the tail, then we have a variable number            of items to iterate over. In the worst case scenario, the index we            want to insert at is just before the tail, meaning we need n-1            iterations through the for loop (found in the else block below),            where n is the number of items in the list. The runtime in this            scenario scales in linear time to the growth of the list, therefore            we disregard the -1 and are left with O(n) runtime. # Check if the given index is out of range and if so raise an error # if we are inserting at the tail, or no previous nodes, then append # if we are inserting at the head, then just perform a prepend # Perform traversal of the nodes # find the node at index, as well as the one just before it # make a new Node for the data to be inserted # insert the new node at the index, and connect it to the rest # increment the size property Insert the given item at the tail of this linked list.             Best and worst case running time: O(1)            In all scenarios, this method runs in constant time. The runtime            complexity of all its statements are independent of the size of the            list. This is because in all the statements we always instantiate a            new Node, check the size of the list, rearrange the pointers of the            head and tail properties (and in some cases the next property of            the tail). Finally, we end by incrementing the value of the size            property. # Create a new node to hold the given item # Check if this linked list is empty # Assign head to new node # Otherwise insert new node after tail # Update tail to new node regardless # Increment the size property Insert the given item at the head of this linked list.             Best and worst case running time: O(1)            Similar to the append method, the runtime of the prepend operation            is independent of the size of the list. In all scenarios, we merely            need to make a new node, rearrange the pointers of the head and            tail, and increment the size of the list. # Create a new node to hold the given item # Check if this linked list is empty # Assign tail to new node # Otherwise insert new node before head # Update head to new node regardless # Increment the size property Return an item from this linked list satisfying the given quality.             Best case running time: O(1) if item is near the head of the list.             Worst case running time: O(n) if item is near the tail of the list            or not present and we need to loop through all n nodes in the list. # Start at the head node # Constant time to assign a variable reference # Loop until the node is None, which is one node too far past the tail # Up to n iterations if we don't exit early # Check if this node's data satisfies the given quality function # Constant time to call quality function # We found data satisfying the quality function, so exit early # Constant time to return data # Skip to the next node # Constant time to reassign a variable # We never found data satisfying quality, but have to return something # Constant time to return None Replace the given old_item in this linked list with given new_item            using the same node, or raise ValueError if old_item is not found.             Best case running time: O(1)            If we are trying to replace an item in a list that's empty, that            has only one item, or the item is present at the head, we have a            constant runtime that is independent of how the size of the list.            In the first scenario mentioned, we would not need any traversal of            the list; and in the other two scenarios, we would need exactly one.             Worst case running time: O(n)            If we are trying to replace an item that is present at the end of            the list, or not present in the list at all, we will need to            traverse through all the items in the list. This operation scales in            linear time, because it grows in direct proportion to the number of            items in the list (since one iteration is required for each item). # Start at the head node # traverse the nodes to find old_item # Check if this node's data matches the old_item # We found the node, now change the data it references # end the method by returning None # Skip to the next node # old_item is not in the list, so raise a ValuError Delete the given item from this linked list, or raise ValueError.             Best case running time: O(1)            If the item is at the head of the list, then we only require one            iteration for traversal, and then we rely on decision strucutres to            appropiately remove the item. The runtime of these operations will            not change based on the length of the list.             Worst case running time: O(n)            If the item at the end of the list, or not present at all, this            method will have to traverse through all the items in the list to            get to it. This causes the runtime of this method to reach O(n)            asymptotically, because it is essentially an implementation of the            linear search algorithm. # Start at the head node # Keep track of the node before the one containing the given item # Create a flag to track if we have found the given item # Loop until we have found the given item or the node is None # Check if the node's data matches the given item # We found data matching the given item, so update found flag # Skip to the next node # Check if we found the given item or we never did and reached the tail # Check if we found a node in the middle of this linked list # Update the previous node to skip around the found node # Unlink the found node from its next node # Check if we found a node at the head # Update head to the next node # Unlink the found node from the next node # Check if we found a node at the tail # Check if there is a node before the found node # Unlink the previous node from the found node # Update tail to the previous node regardless # Decrement the size property # Otherwise raise an error to tell the user that delete has failed Return a generator of the data referenced by the Nodes in this            LinkedList.             Credit to my old LinkedList class in my CS 1.2 repo:            https://github.com/UPstartDeveloper/CS-1.2-Intro-Data-Structures/blob/master/Code/linkedlist.py