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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

I always encourage contributors to participate in the discussion forum for this repository.

If you have a better solution or any queries / discussions related to the Problem of the Day solution, please visit our discussion section. We welcome your input and aim to foster a collaborative learning environment.

If you find this solution helpful, consider supporting us by giving a โญ star to the getlost01/gfg-potd repository.

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gfg-potd's Issues

22 JAN POTD Paths from root with a specified sum

class Solution
{
public:
vector<vector> printPaths(Node *root, int sum)
{
vector<vector> ans;
vector path;

    function<void(Node *, int)> dfs = [&](Node * node, int cur_sum) {
        if(!node)
            return;
        
        cur_sum += node -> key;
        path.push_back(node -> key);
        
        if(cur_sum == sum)
            ans.push_back(path);
        
        dfs(node -> left, cur_sum);
        dfs(node -> right, cur_sum);
        
        cur_sum -= node -> key;
        path.pop_back();
    };
    
    dfs(root, 0);
    
    return ans;
    
}

};

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