CareerCup

Obviously for first task only first t-1 elements in list will survive.

for first part - delete all elements starting from 't'th node to end of list

for second part - delete elements starting from 't'th node to end
now only t-1 elements are remaining, delete the first node from that list - since a traversal of t nodes from 1st node on a list of length t-1 would end at 1st node
From there on follow simple traversal and delete nodes.

Wondering if the last part can be generalized

gevorgk is correct. Move to the t'th element in the list and delete that element and all remaining elements in the list.

MS IDC makes fool of people.
People have to come to office on weekends due to workload and do night outs, no work life balance. They pay 10-20% more make people labour.

Do take the feedback from employees before joining MS.

And work is junk, all junk wor from Redmond is transferred to IDC. Ask any team, whether they design, implement products or just do porting or maintenance or make tools.

For a circular list:

node* RemoveTthNode(int t)
{
   node *curr=head;
   node *temp;
   int count=1;
   //limiting condition(circular list with only 1 element)
   while(curr!=curr->next)
   {
     if (count==t-1)
     {
       temp=curr->next;
       curr->next=temp->next;
       free(temp);
       //Reset count after deleting Tth node
       count=1;
     }
     else
     {
       count++;
       curr=curr->next;
     }
   }

   return curr;
}

2nd Part
Delete every 't' th node till node->next=node

@sal- dude u deserve to be in microsoft or amazon...do u think careercup r bunch of fools to publish d question the way u interpreted......wow :)...wat an answer

guys use 2 pointers...
one will traverse the link list...
other will delete the required node and move by required t postion....
the complexity is o(n)...as d slow pointer is the one which will set the loop condition

guys use 2 pointers...
one will traverse the link list...
other will delete the required node and move by required t postion....
the complexity is o(n)...as d slow pointer is the one which will set the loop condition

This looks like "Lucky Numbers" problem. Please take a look in geeksforgeeks.org

We can solve the problem in O(nlogn)

We can solve the problem in O(nlogn)using tree

^ throw some lights buddy

This is Josephus problem.