Problem Description

In this problem, you are provided with the head of a singly linked list and an integer k. The goal is to return the modified linked list after swapping the values of two specific nodes: the kth node from the beginning and the kth node from the end. The linked list is indexed starting from 1, which means that the first node is considered the 1st node from both the beginning and the end in the case of a single-node list. Note that it's not the nodes themselves that are being swapped, but rather their values.

Intuition

To solve this problem, we think about how we can access the kth node from the beginning and the end of a singly linked list. A singly linked list does not support random access in constant time; meaning we can't directly access an element based on its position without traversing the list.

Given we only need to swap the values and not the nodes themselves, a two-pointer approach is perfect. The first step is to locate the kth node from the beginning which can be done by traversing k-1 nodes from the head. We use a pointer p to keep track of this node.

Finding the kth node from the end is trickier because we don't know the length of the linked list beforehand. However, by using two pointers – fast and slow – and initially set both at the head, we can move fast k-1 steps ahead so that it points to the kth node. Then, we move both fast and slow at the same pace. When fast reaches the last node of the list, slow will be pointing at the kth node from the end. We use another pointer q to mark this node.

Once we have located both nodes to be swapped, p and q, we simply exchange their values and return the (modified) head of the linked list as the result.

Learn more about Linked List and Two Pointers patterns.

Solution Approach

The implementation of the solution follows a two-pointer approach that is commonly used when dealing with linked list problems. The algorithm is simple yet powerful and can be broken down into the following steps:

1. Initialize two pointers fast and slow to reference the head of the linked list.
2. Move the fast pointer k-1 steps ahead. After this loop, fast will be pointing to the kth node from the beginning of the list. We use pointer p to mark this node.
3. Once the fast pointer is in position, we start moving both fast and slow pointers one step at a time until fast is pointing to the last node of the linked list. Now, the slow pointer will be at the kth node from the end of the list. We use another pointer q to mark this node.
4. Swap the values of nodes p and q by exchanging their val property.
5. Finally, return the modified linked list's head.

This is a classical algorithm that does not require additional data structures for its execution and fully utilizes the linked list's sequential access nature. It's efficient because it only requires a single pass to find both the kth node from the beginning and the kth node from the end, resulting in O(n) time complexity, where n is the number of nodes in the linked list.

The only tricky part that needs careful consideration is handling edge cases, such as k being 1 (swapping the first and last elements), k being equal to half the list's length (in the case of an even number of nodes), or the list having k or fewer nodes. However, since the problem only asks to swap values, the provided solution handles all these cases without any additional checks.

Example Walkthrough

Let's walk through an example to illustrate the solution approach. Consider a linked list and an integer k:

Linked List: 1 -> 2 -> 3 -> 4 -> 5
k: 2

Our task is to swap the 2nd node from the beginning with the 2nd node from the end.

1. Initialize two pointers fast and slow to reference the head of the linked list.

2. Move the fast pointer k-1 steps ahead. So, after this step:

   fast: 2 (Second node)
   slow: 1 (Head of the list)

3. Start moving both fast and slow one step at a time until fast points to the last node:

   Step 1:
   fast: 3
   slow: 2
   
   Step 2:
   fast: 4
   slow: 3
   
   Step 3:
   fast: 5 (Last node)
   slow: 4 (This becomes our `k`th node from the end)

4. At this point, we have:

   p (kth from the beginning): Node with value 2
   q (kth from the end): Node with value 4

5. Swap the values of nodes p and q:

   Node `p` value before swap: 2
   Node `q` value before swap: 4
   
   Swap their values.
   
   Node `p` value after swap: 4
   Node `q` value after swap: 2

6. The modified linked list now looks like this:

   Modified Linked List: 1 -> 4 -> 3 -> 2 -> 5

7. Return the head of the modified linked list, which points to the node with value 1.

This example illustrates the power of the two-pointer approach where pointer manipulation allows us to swap the kth nodes' values from the start and end of a singly linked list in O(n) time complexity without any additional space required.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code snippet is O(n), where n is the length of the linked list. This is because the code iterates over the list twice: once to find the k-th node from the beginning, and once more to find the k-th node from the end while simultaneously finding the end of the list.

The space complexity is O(1) since only a constant amount of additional space is used, regardless of the input size. No extra data structures are created that depend on the size of the list; only a fixed number of pointers (fast, slow, and p) are used.

Learn more about how to find time and space complexity quickly using problem constraints.