README.md

GFG Problem Of The Day

Today - 3 July 2024

Que - Remove all occurences of duplicates in a linked list

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

My Approach

• Create an unordered map mp to count the occurrences of each value in the linked list.
• Traverse the linked list using a pointer temp and update the map with the count of each node's data.
• Create a dummy node that points to the head of the linked list. This helps in handling edge cases where the head node might need to be removed.
• Initialize a pointer prev to the dummy node.
• Reset temp to the head of the linked list.
• Traverse the linked list again using temp.
• For each node, check the count of its data in the map:
• If the count is greater than 1, it means the node is a duplicate, so update the next pointer of prev to skip the current node (temp).
• If the count is 1, move the prev pointer to the current node (temp).
• Move the temp pointer to the next node.
• Set new_head to the next of the dummy node.
• Delete the dummy node to free up memory.
• Return new_head, which is the head of the modified linked list with duplicates removed.

Time and Auxiliary Space Complexity

• Time Complexity: O(N), where N is the number of nodes in the linked list.
• Auxiliary Space Complexity: O(N), where N is the number of nodes in the linked list.

Code (C++)

class Solution {
  public:
    Node* removeAllDuplicates(struct Node* head)
    {
        unordered_map<int, int>mp;
        Node* temp=head;
        while (temp)
        {
            mp[temp->data]++;
            temp=temp->next;
        }
        Node* dummy=new Node(0);
        dummy->next=head;
        Node* prev=dummy;
        temp=head;
        while (temp)
        {
            if (mp[temp->data]>1)
                prev->next=temp->next;
            else prev=temp;
            temp=temp->next;
        }
        Node* new_head=dummy->next;
        delete dummy;
        return new_head;
    }
};

Contribution and Support

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