Software Engineer

Explained naive approach first: We can traverse through the Linked List while maintaining a count of nodes, let’s say in variable count , and then traversing for the 2nd time for (n – count) nodes to get to the nth node of the list.

Then went for optimised approach:
1. Take two dummy nodes, who’s next will be pointing to the head.
2.Take another node to store the head, initially it,s a dummy node(start), and the next of the node will be pointing to the head.The reason why we are using this extra dummy node, is because there is an edge case. If the node is equal to the length of the linkedlist, then this slow’s will point to slow’s next→ next. And we can say our dummy start node will be broken, and will be connected to the slow’s next→ next.
3.Start traversing until the fast pointer reaches the nth node.
4.Now start traversing by one step both of the pointers until the fast pointers reach the end.
5. When the traversal is done, just do the deleting part. Make slow pointer’s next to the next of next of the slow pointer to ignore/disconnect the given node.
6.Last, return the next of start.

It is a standard DP problem which can be solved

Think out loud. Ask interviews if you have any questions and doubts.

The idea is to take 12:00 (h = 12, m = 0) as a reference. Following are detailed steps.

1. Calculate the angle made by hour hand with respect to 12:00 in h hours and m minutes. 
2. Calculate the angle made by minute hand with respect to 12:00 in h hours and m minutes. 
3. The difference between the two angles is the angle between the two hands.

The minute hand moves 360 degrees in 60 minute(or 6 degrees in one minute) and hour hand moves 360 degrees in 12 hours(or 0.5 degrees in 1 minute). In h hours and m minutes, the minute hand would move (h*60 + m)*6 and hour hand would move (h*60 + m)*0.5.

Revise basics and practice standard questions asked

Tip 1 : Practice standard questions beforehand.