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DSA Pythonprogramming~5 mins

Delete by Value in Doubly Linked List in DSA Python - Cheat Sheet & Quick Revision

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Recall & Review
beginner
What is a doubly linked list?
A doubly linked list is a chain of nodes where each node has three parts: data, a pointer to the next node, and a pointer to the previous node. This allows moving forward and backward through the list.
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beginner
What does 'delete by value' mean in a doubly linked list?
It means finding the first node that contains a specific value and removing that node from the list, adjusting the pointers of neighboring nodes to keep the list connected.
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intermediate
Which pointers need to be updated when deleting a node in a doubly linked list?
You need to update the 'next' pointer of the previous node and the 'prev' pointer of the next node to bypass the node being deleted.
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intermediate
What happens if the node to delete is the head of the doubly linked list?
The head pointer must be updated to point to the next node, and if that node exists, its 'prev' pointer should be set to None.
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beginner
What should you do if the value to delete is not found in the doubly linked list?
You should leave the list unchanged because there is no node with that value to delete.
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In a doubly linked list, what does each node contain?
AData, next pointer, previous pointer
BData and next pointer only
CData and previous pointer only
DData only
When deleting a node by value, what is the first step?
ADelete the last node
BUpdate the head pointer
CSet all pointers to None
DFind the node with the given value
If the node to delete is in the middle, which pointers are updated?
ANo pointers need updating
BOnly head pointer
CPrevious node's next and next node's prev
DOnly tail pointer
What happens if the node to delete is the only node in the list?
AThe node remains
BThe list becomes empty (head is None)
CThe next pointer is updated
DThe previous pointer is updated
If the value to delete is not found, what should the function do?
ADo nothing and keep the list unchanged
BDelete the head node
CDelete the tail node
DRaise an error
Explain step-by-step how to delete a node by value in a doubly linked list.
Think about how pointers before and after the node change.
You got /5 concepts.
    Describe how pointers change when deleting a middle node in a doubly linked list.
    Focus on the two neighbors of the node.
    You got /3 concepts.

      Practice

      (1/5)
      1. What happens when you delete a node by value in a doubly linked list?
      easy
      A. The list is reversed instead of deleting
      B. All nodes are deleted regardless of value
      C. Only the first node is deleted always
      D. The node with the matching value is removed and links are updated

      Solution

      1. Step 1: Locate the node with the given value

        Traverse the list from head to tail to find the node whose data matches the value to delete.
      2. Step 2: Update links to remove the node

        Adjust the previous node's next pointer and the next node's previous pointer to bypass the node to be deleted.
      3. Final Answer:

        The node with the matching value is removed and links are updated -> Option D
      4. Quick Check:

        Delete by value means remove matching node [OK]
      Hint: Find node, update previous and next pointers [OK]
      Common Mistakes:
      • Not updating both previous and next pointers
      • Deleting all nodes instead of one
      • Ignoring the case when node is head or tail
      2. Which of the following is the correct syntax to delete a node by value in a doubly linked list in Python?
      easy
      A. def delete_by_value(self, value): current = self.head while current: if current.data == value: # update links break current = current.next
      B. def delete_by_value(value): current = head while current: if current.data = value: # update links break current = current.next
      C. def delete_by_value(self, value): current = self.head while current: if current.data == value # update links break current = current.next
      D. def delete_by_value(self, value): current = self.head while current: if current.data == value: # update links continue current = current.next

      Solution

      1. Step 1: Check method signature and loop

        def delete_by_value(self, value): current = self.head while current: if current.data == value: # update links break current = current.next correctly defines a method with self and value, and uses a while loop to traverse nodes.
      2. Step 2: Verify condition and syntax

        def delete_by_value(self, value): current = self.head while current: if current.data == value: # update links break current = current.next uses '==' for comparison and breaks after deletion logic, which is correct syntax and logic.
      3. Final Answer:

        Option A has correct syntax and logic for deletion by value -> Option A
      4. Quick Check:

        Correct method and loop syntax [OK]
      Hint: Use '==' for comparison and break after deletion [OK]
      Common Mistakes:
      • Using '=' instead of '==' in condition
      • Missing colon ':' after if statement
      • Not including self parameter in method
      3. What will be the output of the following code after deleting value 20?
      class Node:
          def __init__(self, data):
              self.data = data
              self.prev = None
              self.next = None
      
      class DoublyLinkedList:
          def __init__(self):
              self.head = None
      
          def append(self, data):
              new_node = Node(data)
              if not self.head:
                  self.head = new_node
                  return
              last = self.head
              while last.next:
                  last = last.next
              last.next = new_node
              new_node.prev = last
      
          def delete_by_value(self, value):
              current = self.head
              while current:
                  if current.data == value:
                      if current.prev:
                          current.prev.next = current.next
                      else:
                          self.head = current.next
                      if current.next:
                          current.next.prev = current.prev
                      break
                  current = current.next
      
          def print_list(self):
              current = self.head
              result = []
              while current:
                  result.append(current.data)
                  current = current.next
              print(result)
      
      dll = DoublyLinkedList()
      dll.append(10)
      dll.append(20)
      dll.append(30)
      dll.delete_by_value(20)
      dll.print_list()
      medium
      A. [10]
      B. [10, 20, 30]
      C. [10, 30]
      D. [20, 30]

      Solution

      1. Step 1: Append nodes 10, 20, 30

        The list becomes 10 <-> 20 <-> 30.
      2. Step 2: Delete node with value 20

        Node 20 is found and removed by updating 10's next to 30 and 30's prev to 10.
      3. Final Answer:

        [10, 30] -> Option C
      4. Quick Check:

        Deleted 20, list is [10, 30] [OK]
      Hint: After deletion, list skips removed value [OK]
      Common Mistakes:
      • Not updating head when deleting first node
      • Not updating prev pointer of next node
      • Printing list before deletion
      4. Identify the error in the following delete_by_value method for a doubly linked list:
      def delete_by_value(self, value):
          current = self.head
          while current:
              if current.data == value:
                  current.prev.next = current.next
                  current.next.prev = current.prev
                  break
              current = current.next
      medium
      A. Does not traverse the list correctly
      B. Does not handle deletion when current is head (current.prev is None)
      C. Uses break instead of continue after deletion
      D. Deletes all nodes with the value instead of one

      Solution

      1. Step 1: Check handling of head node deletion

        If current is head, current.prev is None, so current.prev.next causes error.
      2. Step 2: Confirm missing condition for head update

        Code does not update self.head when deleting the first node, causing attribute error.
      3. Final Answer:

        Does not handle deletion when current is head (current.prev is None) -> Option B
      4. Quick Check:

        Head deletion needs special handling [OK]
      Hint: Check if node is head before updating prev pointer [OK]
      Common Mistakes:
      • Not checking if current.prev is None
      • Assuming all nodes have prev and next
      • Forgetting to update head pointer
      5. Given a doubly linked list with nodes 5 <-> 10 <-> 15 <-> 20, what will be the list after deleting value 5 and then value 20?
      hard
      A. [10, 15]
      B. [5, 10, 15, 20]
      C. [15]
      D. [10, 15, 20]

      Solution

      1. Step 1: Delete node with value 5 (head)

        After deleting 5, head moves to 10, list is 10 <-> 15 <-> 20.
      2. Step 2: Delete node with value 20 (tail)

        After deleting 20, 15 becomes tail, list is 10 <-> 15.
      3. Final Answer:

        [10, 15] -> Option A
      4. Quick Check:

        Deleted head and tail nodes correctly [OK]
      Hint: Delete head and tail carefully, update pointers [OK]
      Common Mistakes:
      • Not updating head after deleting first node
      • Not updating tail after deleting last node
      • Deleting nodes in wrong order
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