28. Delete Middle of Linked List

The problem can be found at the following link: Question Link

My Approach

• Initialize a temporary pointer temp pointing to the head of the linked list and a variable count to 0.

• Traverse the linked list using a while loop until temp becomes nullptr.

• Increment the count variable for each node encountered.

• Check if count is less than or equal to 1, if true, return NULL indicating an empty or single-node list.

• Reset temp to point to the head of the list.

• Calculate the index of the middle node by dividing count by 2 and storing it in mid.

• Traverse the list again until mid - 1 becomes 0, updating temp to point to the middle node.

• Update the next pointer of the node before the middle node to skip the middle node, effectively removing it from the list.

• Return the head of the modified linked list.

Time and Auxiliary Space Complexity

• Time Complexity : O(N)

• Auxiliary Space Complexity : O(1)

Code (C++)

class Solution{
    public:
    Node* deleteMid(Node* head)
    {
        Node* temp=head;
        int count=0;
        while (temp!=nullptr)
        {
            count++;
            temp=temp->next;
        }
        if (count<=1)
            return NULL;
        temp=head;
        int mid=count/2;
        while(mid-1>0)
        {
            temp=temp->next;
            mid--;
        }
        temp->next=temp->next->next;
        return head;
    }
};

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