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Reverse Given Linked List in Groups of K Problem

Given a linked list, reverse every group of k nodes. If there is a reminder (a group of less than k nodes) in the end, reverse that last group too.


Example One

Input: list: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL

k: 3.

Output: 3 -> 2 -> 1 -> 6 -> 5 -> 4 -> NULL

Input list consists of two whole groups of three. In the output list the first three and last three nodes are reversed.


Example Two

Input: list: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL

k: 3.

Output: 3 -> 2 -> 1 -> 6 -> 5 -> 4 -> 8 -> 7 -> NULL

There are two whole groups of three and one partial group (a remainder that consists of just two nodes). Each of the three groups is reversed in the output.


Notes

The function has two parameters: head of the given linked list and k.

Return the head of the linked list after reversing the groups of nodes in it.


Constraints:

● 1

● -2 * 10^9

● 1

● Cannot use more than constant extra space.


Solution

The “constant extra space” constraint rules out using recursion because recursion uses the call stack space (up to O(n) if k=n).

We are forced to use an iterative approach.

iterative_solution.cpp demonstrates how it can be done. The code has comments and is easy to follow.


Time Complexity:

O(n).

Auxiliary Space Used:

O(1).

Space Complexity:

O(n) because of input and output size.


// -------- START --------

// Helper function: Reverse singly linked list in O(length) time and O(1) extra space.
void reverse_linked_list(LinkedListNode *cur)
{
	LinkedListNode *prev = NULL;
	LinkedListNode *next;
	while (cur)
	{
		next = cur->next;
		cur->next = prev;
		prev = cur;
		cur = next;
	}
}

LinkedListNode *reverse_linked_list_in_groups_of_k(LinkedListNode *head, int k)
{
	/*
	Input:
	list: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL
	k: 3
	Output:
	3 -> 2 -> 1 -> 6 -> 5 -> 4 -> 8 -> 7-> NULL
	Groups to be reversed are (1 -> 2 -> 3), (4 -> 5 -> 6) and (7 -> 8).
	We will call reverse_linked_list function when (start = 1 and stop = 3),
	(start = 4 and stop = 6) and (start = 7 and stop = 8). 
	*/
	// Points to previous start node.
	LinkedListNode *prev_of_start = NULL;
	LinkedListNode *start = head;
	LinkedListNode *stop = head;
	int count = 0;
	// Traverse whole linked list.
	while (stop)
	{
		count++;
		/*
		If we have covered k nodes in between start
		and stop (inclusive) or we are at the last node.
		*/
		if (count == k || stop->next == NULL)
		{
			// Points to next node of start. 
			LinkedListNode *next_of_stop = stop->next;
			/*
			We want to reverse start to stop nodes,
			set stop->next = NULL so we know where to stop.
			*/
			stop->next = NULL;
			// Reverse start to stop nodes. 
			reverse_linked_list(start);
			if (prev_of_start == NULL)
			{
				// Head will change when we are reversing the linked list first time. 
				head = stop;
			}
			else
			{
				/*
				We have reversed start to stop nodes, hence now stop will be next node of 
				prev_of_start. 
				*/
				prev_of_start->next = stop;
			}
			/*
			We have reversed start to stop nodes, hence next_of_stop will be next node of start. 
			*/
			start->next = next_of_stop;
			/*
			In the above example, after we have reversed first k nodes list will be:
			3 -> 2 -> 1 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL,
			start will point to 1, next_of_stop will point to 4.
			
			Now we will set start and stop to point at 4.
			And prev_of_start should be previous of 4 that is 1.
			*/
			prev_of_start = start;
			start = next_of_stop;
			stop = next_of_stop;
			// Reset counter. 
			count = 0;
		}
		else
		{
			stop = stop->next;
		}
	}
	return head;
}

// -------- END --------