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DSA Goprogramming~5 mins

Delete by Value in Doubly Linked List in DSA Go - Time & Space Complexity

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Time Complexity: Delete by Value in Doubly Linked List
O(n)
Understanding Time Complexity

When deleting a node by value in a doubly linked list, we want to know how long it takes as the list grows.

We ask: How does the time to find and remove a value change with list size?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


func deleteByValue(head *Node, val int) *Node {
    current := head
    for current != nil {
        if current.data == val {
            if current.prev != nil {
                current.prev.next = current.next
            } else {
                head = current.next
            }
            if current.next != nil {
                current.next.prev = current.prev
            }
            break
        }
        current = current.next
    }
    return head
}
    

This code searches for the first node with the given value and removes it from the doubly linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Traversing the list node by node to find the value.
  • How many times: Up to n times, where n is the number of nodes in the list.
How Execution Grows With Input

As the list grows, the search may take longer because it might check more nodes.

Input Size (n)Approx. Operations
10Up to 10 checks
100Up to 100 checks
1000Up to 1000 checks

Pattern observation: The number of operations grows roughly in direct proportion to the list size.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete by value grows linearly with the number of nodes in the list.

Common Mistake

[X] Wrong: "Deleting by value is always fast because we just remove the node."

[OK] Correct: We must first find the node by checking each one, which can take time proportional to the list size.

Interview Connect

Understanding this helps you explain how searching affects deletion speed in linked lists, a common topic in interviews.

Self-Check

"What if the list was sorted? How would that change the time complexity of deleting by value?"

Practice

(1/5)
1. What happens when you delete a node by value in a doubly linked list?
easy
A. The node with the matching value is removed and links are updated
B. All nodes are deleted regardless of value
C. Only the first node is deleted no matter the value
D. The list is reversed instead of deleting

Solution

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

    Traverse the doubly linked list from head to tail to find the node whose data matches the value to delete.
  2. Step 2: Update the 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, effectively removing it from the list.
  3. Final Answer:

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

    Delete by value means remove matching node and fix links [OK]
Hint: Find node by value, then fix previous and next links [OK]
Common Mistakes:
  • Not updating previous node's next pointer
  • Not updating next node's previous pointer
  • Deleting wrong node when multiple have same value
2. Which of the following Go code snippets correctly deletes a node by value from a doubly linked list?
easy
A. if current.data == value { current.next.next = current.prev.prev }
B. if current.data == value { current.next = current.prev; current.prev = current.next }
C. if current.data == value { current = nil }
D. if current.data == value { current.prev.next = current.next; current.next.prev = current.prev }

Solution

  1. Step 1: Check node data against value

    The condition current.data == value correctly identifies the node to delete.
  2. Step 2: Update links correctly

    Setting current.prev.next = current.next and current.next.prev = current.prev properly removes the node by linking its neighbors.
  3. Final Answer:

    if current.data == value { current.prev.next = current.next; current.next.prev = current.prev } -> Option D
  4. Quick Check:

    Correct link updates remove node [OK]
Hint: Update prev.next and next.prev to remove node [OK]
Common Mistakes:
  • Swapping prev and next pointers incorrectly
  • Setting current to nil without fixing links
  • Accessing next or prev without nil checks causing panic
3. Given the doubly linked list: 1 <-> 2 <-> 3 <-> 4 <-> 5, what is the list after deleting the node with value 3?
medium
A. 1 <-> 2 <-> 4 <-> 3 <-> 5
B. 1 <-> 2 <-> 4 <-> 5
C. 2 <-> 3 <-> 4 <-> 5
D. 1 <-> 2 <-> 3 <-> 5

Solution

  1. Step 1: Identify node with value 3

    Traverse the list to find the node containing 3, which is between nodes 2 and 4.
  2. Step 2: Remove node 3 and update links

    Set node 2's next pointer to node 4 and node 4's previous pointer to node 2, skipping node 3.
  3. Final Answer:

    1 <-> 2 <-> 4 <-> 5 -> Option B
  4. Quick Check:

    Node 3 removed, neighbors linked [OK]
Hint: Remove node and link neighbors directly [OK]
Common Mistakes:
  • Leaving node 3 linked causing broken list
  • Deleting wrong node by value
  • Not updating both prev and next pointers
4. Consider this Go code snippet to delete a node by value in a doubly linked list:
func deleteByValue(head *Node, value int) *Node {
    current := head
    for current != nil {
        if current.data == value {
            if current.prev != nil {
                current.prev.next = current.next
            } else {
                head = current.next
            }
            if current.next != nil {
                current.next.prev = current.prev
            }
            break
        }
        current = current.next
    }
    return head
}
What is the bug in this code?
medium
A. It does not handle deleting the last node correctly
B. It does not update head when deleting the first node
C. It breaks after deleting only the first matching node
D. It does not check for nil pointers before accessing prev or next

Solution

  1. Step 1: Analyze loop and break behavior

    The loop breaks immediately after deleting the first node with matching value, so only one node is deleted even if multiple nodes have the same value.
  2. Step 2: Understand implications for multiple matches

    If the list has multiple nodes with the same value, only the first is removed, which may not be the intended behavior if all matches should be deleted.
  3. Final Answer:

    It breaks after deleting only the first matching node -> Option C
  4. Quick Check:

    Break stops after first delete [OK]
Hint: Check if loop breaks too early when deleting multiple nodes [OK]
Common Mistakes:
  • Ignoring multiple nodes with same value
  • Not updating head when first node deleted
  • Not checking nil before pointer access
5. You have a doubly linked list with nodes: 10 <-> 20 <-> 30 <-> 20 <-> 40. You want to delete all nodes with value 20. Which approach correctly deletes all such nodes?
hard
A. Traverse the list and delete nodes with value 20, continuing without breaking the loop
B. Delete the first node with value 20 and then stop traversal
C. Delete nodes with value 20 only if they are at the head or tail
D. Replace nodes with value 20 with a new node containing 0

Solution

  1. Step 1: Traverse entire list without breaking

    To delete all nodes with value 20, you must check every node and not stop after the first deletion.
  2. Step 2: Delete matching nodes and update links carefully

    For each node with value 20, update the previous and next pointers to remove it, then continue traversal to find others.
  3. Final Answer:

    Traverse the list and delete nodes with value 20, continuing without breaking the loop -> Option A
  4. Quick Check:

    Delete all matches by full traversal [OK]
Hint: Do not break loop; delete all matching nodes during traversal [OK]
Common Mistakes:
  • Breaking loop after first deletion
  • Not updating links properly causing broken list
  • Only deleting head or tail nodes
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