Delete Nth node from end of Linked List
You have been given a singly Linked List of 'N' nodes with integer data and an integer 'K'. Your task is to remove the Kth node from the end of the given Linked List.

For example:

The given linked list is 1 - 2 - 3 - 4- 'NULL'. and 'K' is 2 After removing the second node from the end, the linked list become 1-2-4-'NULL'

Follow Up:

Can you solve this without finding the length of the linked list and using O(1) extra space?

Input format :

The first line of input contains an integer 'T' representing the number of test cases or queries to be processed. Then the test case follows. The first line of each test case contains a single integer 'K', representing the index(1 based indexing) of a node from the last to be deleted. The second line of each test case contains the elements of the singly linked list separated by a single space and terminated by -1. Hence, -1 would never be a list element.

Output format :

For each test case, print a single line that contains the updated linked list in a linear fashion. A single space will separate all the list data and -1 will indicate the end of the list. Print output of each test case in a separate line.

Note :

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints :

1

Question

Delete Nth node from end of Linked List

You have been given a singly Linked List of 'N' nodes with integer data and an integer 'K'. Your task is to remove the Kth node from the end of the given Linked List.

For example:

The given linked list is 1 -> 2 -> 3 -> 4-> 'NULL'. and 'K' is 2
After removing the second node from the end, the linked list become 1->2->4->'NULL'

Follow Up:

Can you solve this without finding the length of the linked list and using O(1) extra space?

Input format :

The first line of input contains an integer 'T' representing the number of test cases or queries to be processed. Then the test case follows.

The first line of each test case contains a single integer 'K', representing the index(1 based indexing) of a node from the last to be deleted.

The second line of each test case contains the elements of the singly linked list separated by a single space and terminated by -1. Hence, -1 would never be a list element.

Output format :

For each test case, print a single line that contains the updated linked list in a linear fashion. A single space will separate all the list data and -1 will indicate the end of the list.

Print output of each test case in a separate line.

Note :

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints :

1 <= 'T' <= 5
0 <= 'N' <= 10 ^ 5
0 <= 'K' <= 'N'
1 <= 'DATA' <= 10 ^ 9 and 'DATA' != -1

Time Limit: 1 sec.

Accepted Answer

Intuition:
Lets K be the total nodes in the linked list.
Observation : The Nth node from the end is (K-N+1)th node from the beginning.
So the problem simplifies down to that we have to find (K-N+1)th node from the beginning.

One way of doing it is to find the length (K) of the linked list in one pass and then in the second pass move (K-N+1) step from the beginning to reach the Nth node from the end.
To do it in one pass. Let’s take the first pointer and move N step from the beginning. Now the first pointer is (K-N+1) steps away from the last node, which is the same number of steps the second pointer require to move from the beginning to reach the Nth node from the end.

Approach:
Take two pointers; the first will point to the head of the linked list and the second will point to the Nth node from the beginning.
Now keep incrementing both the pointers by one at the same time until the second is pointing to the last node of the linked list.
After the operations from the previous step, the first pointer should point to the Nth node from the end now. So, delete the node the first pointer is pointing to.

Accepted Answer

Using Lists

The naive solution is to process all the nodes from the front side of the linked list and keep adding a node to the list. Now we can easily remove the Kth node from the end of the list by simply replacing the next pointer of the ('LENGTH' - 'K' - 1)th node (0-based indexing from start) of the list with the ('LENGTH' - 'K' + 1)th node. This way we can remove the Kth node from the end of the linked list.

Space Complexity: O(n)Explanation:

O(N), where N is the number of nodes in a linked list.

In the worst case, we will be storing all nodes of the linked list in a new list.

Time Complexity: O(n)Explanation:

O(N), where N is the number of nodes in the linked list.

In the worst-case, we will be iterating the whole linked list. Hence, the overall complexity will be O(N).

Accepted Answer

By Finding Length of the Linked List

Find the length of Linked List 'L'
Check if 'L' = 'K' then remove the head by assigning head to head’s next node.
Else start iterating through the linked list until it comes to ('L' - 'K' -1 )th(0 based indexing from start) node. We will remove the ('L' - 'K')th node by relinking the next pointer of ('L' - 'K' - 1)th node to the ('L' - 'K' + 1)th node.

Space Complexity: O(1)Explanation:

O(1).

In the worst case, only a constant extra space is required.

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the linked list.

In the worst case, we will be traversing the whole linked list. Hence, the overall complexity is O(N).

Accepted Answer

Using Slow and Fast Pointers

We can remove the required node without finding the length of the given linked list by using two pointers 'SLOW' and 'FAST', which are 'K' nodes apart from each other.

Initially, the 'FAST' pointer advances the list by 'K' nodes from the beginning and the 'SLOW' is a pointer to the head of the linked list.
Now both pointers are exactly separated by 'K' distance from each other. We will maintain a constant gap by advancing both pointers together until the 'FAST' pointer reaches the last node.
When the 'FAST' pointer is at the last node then the 'SLOW' pointer will be at ('K' + 1)th node from the end of the linked list.
At last, we will set the next of 'SLOW' pointer to its next of next node.

Space Complexity: O(1)Explanation:

O(1).

In the worst case, only a constant extra space is required.

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the linked list.

In the worst case, we will be traversing the whole linked list. Hence, the overall complexity is O(N).

You have been given a singly Linked List of 'N' nodes with integer data and an integer 'K'. Your task is to remove the Kth node from the end of the given Linked List.

For example:

Follow Up:

Input format :

Output format :

Note :

Constraints :

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