Last Updated on April 21, 2022 by Ria Pathak
Introduction
The linked list is one of the most important concepts to know while preparing for interviews. Having a good grasp of a linked list can be a huge plus point in coding interviews.
Problem Statement
We will be given a circular linked list and a position, and we need to delete the element present at that position in the list.
Note: we need to follow 0 based indexing while deleting the nodes
Problem Statement Understanding
Let’s try to understand the problem with the help of an example.
If the linked list given to us is:
Now, if position = 0.
- Then, we need to delete the first node of the list, and after deletion, the list will now look like this:
If the position = 2.
- Then, we need to delete the third node, and the list after deletion will look like this:
At this point, we have understood the problem statement. Now we will try to formulate an approach for this problem.
Before moving to the approach section, try to think about how you can approach this problem.
- If stuck, no problem, we will thoroughly see how we can approach this problem in the next section.
Let’s move to the approach section.
Approach
We will divide our problem into three sections:
1) When position = 0:
a) It means that we need to delete the first node of the circular linked list.
b) In this case, we need to reach the last node and then connect it with the second node.
c) After that, we will update the head pointer for the list.
2) When position = length of the list (we need to delete the last node):
a) In this case, we need to reach the last second node and connect it with the first node.
3) When we need to delete a middle node:
a) We need to keep track of node position while iterating the list.
b) when we find the node at the specified position, we need to connect its previous node with its next node.
Let’s move to the algorithm section to see the above approach in more depth.
Algorithm
1) First, calculate the length of the list and save it in len variable.
2) Initialize a variable count by 1.
3) If the head is NULL, then it means that the list is empty, so return from the function.
4) If the position is less than 0 or greater than len, then return from the function.
5) If position == 0 then, call deleteFirstNode function:
- Initialize curr and left with the head of the list.
- Return from the function if the head is NULL.
- Make the head NULL and return from the function, if left->next is left (this is the case when there is only 1 node in the list).
- Run a while loop till left->next is not equal to head.
- Inside the loop, advance left by one node and update curr with the next node of left.
- Now, when the loop is executed completely, connect the next pointer of left to the next node of curr.
- Update the head node by left->next.
- Delete the old head node, i.e., curr.
6) If position == (len-1) then, call deleteLastNode function: - Initialize the curr pointer with head and left with NULL.
- Return from the function if the head is NULL.
- Make the head NULL and return from the function, if curr->next is curr (this is the case when there is only 1 node in the list).
- Run a while loop till curr->next is not equal to head.
- Inside the loop, save the value of curr in left and advance curr by one node.
- Now, when the loop is executed completely, connect the next pointer of left to the next node of curr.
- Update the head node by left->next.
- Delete the old head node, i.e., curr.
7) Run a while loop till len is positive. - If position is equal to count then,
- Connect previous node’s next pointer to current node’s next pointer.
- Delete the curr pointer and return from the function.
- Advance previous by one node and update curr to the node next to previous.
- Decrement len and increment count by one.
8) Return from the function once while loop breaks.
Dry Run
Code Implementation
#includeusing namespace std; class Node { public: int data; Node* next; Node(int d) { data = d; next = NULL; } }; // This function will print data of // all the nodes present in the list void Display(Node* head) { Node* temp = head; // In case of an empty list: if (head == NULL) { printf("\nDisplay List is empty\n"); return; } // iterate the complete list else { do { cout<<(temp->data); temp = temp->next; cout<<" "; } while (temp != head); } cout<<'\n'; } // This function calculates the // number of nodes in the list int Length(Node* head) { Node* temp = head; int len = 0; // in case of an empty list, return 0 if (head == NULL) { return 0; } // iterate the complete list else { do { temp = temp->next; len++; } while (temp != head); } return len; } // This node deletes the first node // of the list void deleteFirstNode(Node** head) { Node *left = *head, *curr = *head; // if the list is empty if (*head == NULL) { printf("\nList is empty\n"); return; } // In case, the list has a single node if (left->next == left) { *head = NULL; return; } // iterate the complete list while (left->next != *head) { left = left->next; curr = left->next; } // left will be pointing to the last node // and curr will be the head of the list // so, we will connect left with the curr->next left->next = curr->next; // make second node as head node *head = left->next; free(curr); return; } // This function deletes the last // node of the list void deleteLastNode(Node** head) { Node *curr = *head, *left = NULL; // if the list is empty if (*head == NULL) { printf("\nList is empty\n"); return; } // In case, the list has a single node if (curr->next == curr) { *head = NULL; return; } //iterate the list while (curr->next != *head) { left = curr; curr = curr->next; } left->next = curr->next; *head = left->next; free(curr); return; } void deleteAtPosition(Node** head, int position) { // find length of the list int len = Length(*head); int count = 1; Node *previous = *head, *curr = *head; // check if the list is empty if (*head == NULL) { printf("\nDelete Last List is empty\n"); return; } // check if the given position is out of // bound of the list if (position >= len || position < 0) { printf("\nposition is not Found\n"); return; } // first node deletion if (position == 0) { deleteFirstNode(head); return; } // last node deletion else if(position == len-1){ deleteLastNode(head); return; } // iterate the complete list while (len > 0) { // delete the node when position is found if (position == count) { previous->next = curr->next; free(curr); return; } previous = previous->next; curr = previous->next; len--; count++; } return; } int main() { Node *head = new Node(3); head->next = new Node(7); head->next->next = new Node(2); head->next->next->next = new Node(5); head->next->next->next->next = head; cout<<"Original list "; Display(head); cout<<"After deleting third node "; deleteAtPosition(&head,2); Display(head); cout<<"After deleting last node "; deleteAtPosition(&head,Length(head)-1); Display(head); cout<<"After deleting first node "; deleteAtPosition(&head,0); Display(head); }
#include#include struct Node { int data; struct Node* next; }; // This function will print data of // all the nodes present in the list void Display(struct Node* head) { struct Node* temp = head; // In case of an empty list: if (head == NULL) { printf("\nDisplay List is empty\n"); return; } // iterate the complete list else { do { printf("%d",temp->data); temp = temp->next; printf(" "); } while (temp != head); } printf("\n"); } // This function calculates the // number of nodes in the list int Length(struct Node* head) { struct Node* temp = head; int len = 0; // in case of an empty list, return 0 if (head == NULL) { return 0; } // iterate the complete list else { do { temp = temp->next; len++; } while (temp != head); } return len; } // This node deletes the first node // of the list void deleteFirstNode(struct Node** head) { struct Node* left = *head; struct Node *curr = *head; // if the list is empty if (*head == NULL) { printf("\nList is empty\n"); return; } // In case, the list has a single node if (left->next == left) { *head = NULL; return; } // iterate the complete list while (left->next != *head) { left = left->next; curr = left->next; } // left will be pointing to the last node // and curr will be the head of the list // so, we will connect left with the curr->next left->next = curr->next; // make second node as head node *head = left->next; free(curr); return; } // This function deletes the last // node of the list void deleteLastNode(struct Node** head) { struct Node *curr = *head, *left = NULL; // if the list is empty if (*head == NULL) { printf("\nList is empty\n"); return; } // In case, the list has a single node if (curr->next == curr) { *head = NULL; return; } //iterate the list while (curr->next != *head) { left = curr; curr = curr->next; } left->next = curr->next; *head = left->next; free(curr); return; } void deleteAtPosition(struct Node** head, int position) { // find length of the list int len = Length(*head); int count = 1; struct Node *previous = *head, *curr = *head; // check if the list is empty if (*head == NULL) { printf("\nDelete Last List is empty\n"); return; } // check if the given position is out of // bound of the list if (position >= len || position < 0) { printf("\nposition is not Found\n"); return; } // first node deletion if (position == 0) { deleteFirstNode(head); return; } // last node deletion else if(position == len-1){ deleteLastNode(head); return; } // iterate the complete list while (len > 0) { // delete the node when position is found if (position == count) { previous->next = curr->next; free(curr); return; } previous = previous->next; curr = previous->next; len--; count++; } return; } void push(struct Node** head_ref, int data) { struct Node* ptr1 = (struct Node*)malloc(sizeof(struct Node)); ptr1->data = data; ptr1->next = *head_ref; if (*head_ref != NULL) { struct Node* temp = *head_ref; while (temp->next != *head_ref) temp = temp->next; temp->next = ptr1; } else ptr1->next = ptr1; /*For the first node */ *head_ref = ptr1; } int main() { struct Node* head = NULL; push(&head, 2); push(&head, 5); push(&head, 7); push(&head, 8); push(&head, 10); printf("Original list "); Display(head); printf("After deleting third node "); deleteAtPosition(&head,2); Display(head); printf("After deleting last node "); deleteAtPosition(&head,Length(head)-1); Display(head); printf("After deleting first node "); deleteAtPosition(&head,0); Display(head); }
class DeletetionAtPos
{
// structure for a node
static class Node
{
int data;
Node next;
};
// Function to insert a node at the end of
// a Circular linked list
static Node Insert(Node head, int data)
{
Node current = head;
// Create a new node
Node newNode = new Node();
// insert data into newly created node
newNode.data = data;
// check list is empty
// if not have any node then
// make first node it
if (head == null)
{
newNode.next = newNode;
head = newNode;
return head;
}
// if list have already some node
else
{
// move first node to last node
while (current.next != head)
{
current = current.next;
}
// put first or head node address
// in new node link
newNode.next = head;
// put new node address into last
// node link(next)
current.next = newNode;
}
return head;
}
// Function print data of list
static void Display( Node head)
{
Node current = head;
// if list is empty, simply show message
if (head == null)
{
System.out.printf("\nDisplay List is empty\n");
return;
}
// traverse first to last node
else
{
do
{
System.out.printf("%d ", current.data);
current = current.next;
} while (current != head);
}
}
// Function return number of nodes present in list
static int Length(Node head)
{
Node current = head;
int count = 0;
// if list is empty
// simply return length zero
if (head == null)
{
return 0;
}
// traverse first to last node
else
{
do
{
current = current.next;
count++;
} while (current != head);
}
return count;
}
// Function delete First node of
// Circular Linked List
static Node DeleteFirst(Node head)
{
Node previous = head, next = head;
// check list have any node
// if not then return
if (head == null)
{
System.out.printf("\nList is empty\n");
return null;
}
// check list have single node
// if yes then delete it and return
if (previous.next == previous)
{
head = null;
return null;
}
// traverse second to first
while (previous.next != head)
{
previous = previous.next;
next = previous.next;
}
// now previous is last node and
// next is first node of list
// first node(next) link address
// put in last node(previous) link
previous.next = next.next;
// make second node as head node
head = previous.next;
return head;
}
// Function to delete last node of
// Circular Linked List
static Node DeleteLast(Node head)
{
Node current = head, previous=null;
// check if list doesn't have any node
// if not then return
if (head == null)
{
System.out.printf("\nList is empty\n");
return null;
}
// check if list have single node
// if yes then delete it and return
if (current.next == current)
{
head = null;
return null;
}
// move first node to last
// previous
while (current.next != head)
{
previous = current;
current = current.next;
}
previous.next = current.next;
head = previous.next;
return head;
}
// Function delete node at a given position
// of Circular Linked List
static Node DeleteAtPosition( Node head, int index)
{
// Find length of list
int len = Length(head);
int count = 1;
Node previous = head, next = head;
// check list have any node
// if not then return
if (head == null)
{
System.out.printf("\nDelete Last List is empty\n");
return null;
}
// given index is in list or not
if (index >= len || index < 0)
{
System.out.printf("\nIndex is not Found\n");
return null;
}
// delete first node
if (index == 0)
{
head = DeleteFirst(head);
return head;
}
// traverse first to last node
while (len > 0)
{
// if index found delete that node
if (index == count)
{
previous.next = next.next;
return head;
}
previous = previous.next;
next = previous.next;
len--;
count++;
}
return head;
}
// Driver Code
public static void main(String args[])
{
Node head = null;
head = Insert(head, 99);
head = Insert(head, 11);
head = Insert(head, 22);
head = Insert(head, 33);
head = Insert(head, 44);
head = Insert(head, 55);
head = Insert(head, 66);
// Deleting Node at position
System.out.printf("Initial List: ");
Display(head);
System.out.printf("\nAfter Deleting node at index 4: ");
head = DeleteAtPosition(head, 4);
Display(head);
// Deleting first Node
System.out.printf("\n\nInitial List: ");
Display(head);
System.out.printf("\nAfter Deleting first node: ");
head = DeleteFirst(head);
Display(head);
// Deleting last Node
System.out.printf("\n\nInitial List: ");
Display(head);
System.out.printf("\nAfter Deleting last node: ");
head = DeleteLast(head);
Display(head);
}
}
class Node:
def __init__(self, new_data):
self.data = new_data
self.next = None
self.prev = None
# This function to insert a node at the end of a list
def Insert(head, data):
current = head
newNode = Node(0)
if (newNode == None):
print("\nMemory Error\n")
return None
newNode.data = data
if (head == None) :
newNode.next = newNode
head = newNode
return head
else:
while (current.next != head):
current = current.next
newNode.next = head
current.next = newNode
return head
# This function will print data of all the nodes present in the list
def Display(head):
current = head
# In case of an empty list:
if (head == None):
print("\nDisplay List is empty\n")
return
# iterate the complete list
else:
while(True):
print( current.data,end=" ")
current = current.next
if (current == head):
break
# This function calculates the number of nodes in the list
def Length(head):
current = head
count = 0
# in case of an empty list, return 0
if (head == None):
return 0
# iterate the complete list
else:
while(True):
current = current.next
count = count + 1
if (current == head):
break
return count
# This node deletes the first node of the list
def deleteFirstNode(head):
previous = head
next = head
if (head == None) :
print("\nList is empty")
return None
if (previous.next == previous) :
head = None
return None
while (previous.next != head):
previous = previous.next
next = previous.next
previous.next = next.next
head = previous.next
return head
# This function deletes the last node of the list
def DeleteLast(head):
current = head
temp = head
previous = None
if (head == None):
print("\nList is empty")
return None
if (current.next == current) :
head = None
return None
while (current.next != head):
previous = current
current = current.next
previous.next = current.next
head = previous.next
return head
# This function delete node at a given position of List
def DeleteAtPosition(head, index):
# Find length of list
len = Length(head)
count = 1
previous = head
next = head
# check if the list is empty
if (head == None):
print("\nDelete Last List is empty")
return None
# check if the given position is out of bound of the list
if (index >= len or index < 0) :
print("\nPosition is not Found")
return None
# first node deletion
if (index == 0) :
head = deleteFirstNode(head)
return head
# iterate the complete list
while (len > 0):
# delete the node when position is found
if (index == count):
previous.next = next.next
return head
previous = previous.next
next = previous.next
len = len - 1
count = count + 1
return head
head = None
head = Insert(head, 3)
head = Insert(head, 7)
head = Insert(head, 2)
head = Insert(head, 5)
print("Original List: ")
Display(head)
print("\nAfter deleting node at index 3: ")
head = DeleteAtPosition(head, 2)
Display(head)
print("\nAfter deleting last node: ")
head = DeleteLast(head)
Display(head)
print("\nAfter Deleting first node: ")
head = deleteFirstNode(head)
Display(head)
Output
Original list 3 7 2 5
After deleting third node 3 7 5
After deleting last node 3 7
After deleting first node 7
Time Complexity: O(n), where n is the number of nodes in the list
[forminator_quiz id=”5474″]
So, in this blog, we have tried to explain how you can delete nodes at different positions in a circular list in the most optimal way. If you want to practice more questions on linked lists, feel free to solve them at Prepbytes (Linked Lists).