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Ainur - C16 Pine #55

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172 changes: 132 additions & 40 deletions linked_list/linked_list.py
Original file line number Diff line number Diff line change
Expand Up @@ -13,67 +13,125 @@ def __init__(self):

# returns the value in the first node
# returns None if the list is empty
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(1)
# Space Complexity: O(1)
def get_first(self):

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pass
if self.head == None:
return None
return self.head.value


# method to add a new node with the specific data value in the linked list
# insert the new node at the beginning of the linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(1)
# Space Complexity: O(1)
def add_first(self, value):

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pass
self.head = Node(value, self.head)

# method to find if the linked list contains a node with specified value
# returns true if found, false otherwise
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def search(self, value):

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pass
current = self.head
while current != None:
if current.value == value:
return True
current = current.next
return False

# method that returns the length of the singly linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def length(self):

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pass
count = 0
current = self.head
while current != None:
count += 1
current = current.next
return count


# method that returns the value at a given index in the linked list
# index count starts at 0
# returns None if there are fewer nodes in the linked list than the index value
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def get_at_index(self, index):

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pass
current = self.head
current_index = 0

while current != None:
if current_index == index:
return current.value
current_index += 1
current = current.next
return None

# method that returns the value of the last node in the linked list
# returns None if the linked list is empty
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n) or O(1)
# Space Complexity: O(1)
def get_last(self):

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✨ However, when would this implementation be O(1)?

pass
if self.head == None:
return None
current = self.head
while current.next != None:
current = current.next
return current.value

# method that inserts a given value as a new last node in the linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n) or O(1)
# Space Complexity: O(1)
def add_last(self, value):

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✨ In which cases would the time complexity be O(1)?

pass
new_node = Node(value)
if self.head == None:
self.head = new_node
return new_node
current = self.head

while current.next != None:
current = current.next
current.next = new_node

# method to return the max value in the linked list
# returns the data value and not the node
def find_max(self):
pass
if self.head == None:

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return None
current = self.head
current_max = self.head.value


while current != None:
if current_max < current.value:
current_max = current.value
current = current.next
return current_max

# method to delete the first node found with specified value
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def delete(self, value):

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✨ Good on space and time complexity, however see comments below ⬇️

pass
current = self.head
prev = None
while current != None:
if current.value == value:
if current == self.head:
self.head = self.head.next
current = self.head

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🤔 At this point, I would recommend exiting the function. Otherwise, you will continue iterating through and end up deleting _all_nodes found with the specified value. This method only asks you to delete the first.

Suggested change
current = self.head
return

else:
prev.next = current.next
current = current.next

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Again, once you have found the value, you can stop looping

Suggested change
current = current.next
return

else:
prev = current
current = current.next

return self.head

# method to print all the values in the linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def visit(self):

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helper_list = []
current = self.head
Expand All @@ -86,32 +144,66 @@ def visit(self):

# method to reverse the singly linked list
# note: the nodes should be moved and not just the values in the nodes
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def reverse(self):

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pass


current = self.head
prev = None
while current != None:
next = current.next
current.next = prev
prev = current
current = next
self.head = prev
## Advanced/ Exercises
# returns the value at the middle element in the singly linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def find_middle_value(self):

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pass
slow = self.head
fast = self.head
while fast != None and fast.next != None:
fast = fast.next.next
slow = slow.next
return slow.value

# find the nth node from the end and return its value
# assume indexing starts at 0 while counting to n
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def find_nth_from_end(self, n):

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✨ Nice!

pass
if self.head == None:
return None
temp = self.head
length = 0
while temp != None:
current = temp.next
length += 1

if n > length:
return None
current = self.head
for i in range(0, length-n):
current = current.next
return current.value


# checks if the linked list has a cycle. A cycle exists if any node in the
# linked list links to a node already visited.
# returns true if a cycle is found, false otherwise.
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def has_cycle(self):

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pass
slow = self.head
fast = self.head

while fast != None and fast.next != None:
slow = slow.next
fast = fast.next.next
if slow == fast:
return True
return False

# Helper method for tests
# Creates a cycle in the linked list for testing purposes
Expand Down
12 changes: 6 additions & 6 deletions tests/linked_list_test.py
Original file line number Diff line number Diff line change
Expand Up @@ -197,7 +197,7 @@ def test_reverse_will_reverse_five_element_list(list):
for i in range(0, 5):
assert list.get_at_index(i) == i

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_find_middle_value_returns_middle_element_of_five_element_list(list):
list.add_first(10)
list.add_first(30)
Expand All @@ -206,7 +206,7 @@ def test_find_middle_value_returns_middle_element_of_five_element_list(list):
list.add_first(20)
assert list.find_middle_value() == 50

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_find_middle_value_returns_element_at_index_two_of_six_element_list(list):
list.add_first(10)
list.add_first(30)
Expand All @@ -216,11 +216,11 @@ def test_find_middle_value_returns_element_at_index_two_of_six_element_list(list
list.add_first(100)
assert list.find_middle_value() == 60

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_nth_from_n_when_list_is_empty(list):
assert list.find_nth_from_end(3) == None

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_find_nth_from_n_when_length_less_than_n(list):
list.add_first(5)
list.add_first(4)
Expand All @@ -230,7 +230,7 @@ def test_find_nth_from_n_when_length_less_than_n(list):

assert list.find_nth_from_end(6) == None

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_find_nth_from_n(list):
list.add_first(1)
list.add_first(2)
Expand All @@ -243,7 +243,7 @@ def test_find_nth_from_n(list):
assert list.find_nth_from_end(3) == 4
assert list.find_nth_from_end(4) == None

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_has_cycle(list):
assert list.has_cycle() == False

Expand Down