Reverse Nodes In K Group Leetcode Problem 25 [Python Solution]

In this guide, we will eat up Reverse Nodes In K Group LeetCode problem, though it might still be that hard Linked List puzzle to you now, not after this guide.

We’ll provide a detailed Python solution, discuss the problem overview, constraints, and reasoning behind our approach.

So, let’s dive in!

Problem Overview

The problem statement goes as follows: Given the head of a linked list, reverse the nodes of the list k at a time and return the modified list.

Here’s what that means:

k is a positive integer and should be less than or equal to the length of the linked list.
If the number of nodes is not a multiple of k, the left-out nodes at the end should remain as they are.
You may not alter the values in the list’s nodes, only the nodes themselves can be changed.

Let’s break this down further with an example.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

In this example, we take the input linked list, which has five nodes, and we reverse it in groups of two.

So, the first group [1,2] becomes [2,1], the second group [3,4] becomes [4,3], and the single node 5 remains unchanged.

The result is [2,1,4,3,5].

Understanding Constraints

Before we dive into the solution, let’s understand the constraints of this problem:

The number of nodes in the list is denoted as ‘n’.
The value of ‘k’ must satisfy: 1 <= k <= n <= 5000.
The value of ‘Node.val’ (the value of each node) is within the range 0 <= Node.val <= 1000.

Now that we have a clear understanding of the problem, let’s look at an efficient Python solution.

Efficient Python Code Solution

Here’s the Python code to solve the Reverse Nodes In K Group problem efficiently:

View on GitHub ➹

def reverseKGroup(self, head: ListNode, k: int) -&gt; ListNode:
    dummy = ListNode(0, head)
    groupPrev = dummy

    while True:
        kth = self.getKth(groupPrev, k)
        if not kth:
            break
        groupNext = kth.next

        # Reverse group
        prev, curr = kth.next, groupPrev.next
        while curr != groupNext:
            tmp = curr.next
            curr.next = prev
            prev = curr
            curr = tmp

        tmp = groupPrev.next
        groupPrev.next = kth
        groupPrev = tmp
    return dummy.next

def getKth(self, curr, k):
    while curr and k &gt; 0:
        curr = curr.next
        k -= 1
    return curr

Now, let’s break down this solution step by step.

1. Initialization

We start by initializing a dummy node.

This is a common technique in linked list problems to avoid dealing with special cases, such as the need to change the head of the list.

The dummy node allows us to simplify the code.

2. Reversing in K Groups

We enter a while True loop, indicating that we will keep processing the linked list until a specific condition is met.

We find the kth node using the getKth function.

This function is used to identify the K-th node in the group we want to reverse.

If it doesn’t exist (i.e., the group is too small to reverse), we break out of the loop.

We also save the groupNext node, which is the node right after the group we are going to reverse.

3. Reversing the Group

The actual reversal of the group happens in the next steps.

We use the prev and curr pointers to traverse the group and reverse the direction of the next pointers.

prev points to the node before the current node (curr).
curr points to the current node.
We use a tmp variable to temporarily store the next pointer of the curr node.
We then update the next pointer of the curr node to point to the prev node.
prev is updated to curr.
curr is updated to tmp.

This process effectively reverses the direction of next pointers for the nodes in the group.

4. Connecting the Reversed Group

After reversing the group, we need to connect it back to the rest of the linked list.

We do this by updating the next pointer of the node right before the group (stored in groupPrev) to point to the kth node (which is now the last node in the reversed group).

This step ensures that the reversed group remains a part of the overall linked list.

5. Moving to the Next Group

Before moving to the next group, we update the groupPrev pointer to the original first node of the group.

This prepares it for the next iteration of the loop, where it will become the node right before the next group.

6. Returning the Result

Finally, we return the modified linked list.

Since we used a dummy node at the beginning, we return dummy.next, which is the new head of the reversed list.

This efficient algorithm allows us to reverse the linked list in groups of size k while handling various edge cases.

Time and Space Complexity

Let’s analyze the time and space complexity of this solution:

Time Complexity: The time complexity is O(n), where n is the number of nodes in the linked list.

We reverse each node only once.

Space Complexity: The space complexity is O(1) because we use a constant amount of extra space to store variables and pointers.

The space required does not depend on the size of the input.

Reasoning Behind Our Approach

The reasoning behind this approach is to efficiently reverse the linked list in k-group increments while keeping track of the necessary pointers.

By using a dummy node, we simplify the handling of edge cases, including the first group and any remaining nodes at the end.

The core of the algorithm is reversing a group of nodes using a pair of pointers (prev and curr).

We also ensure that the reversed group is correctly connected to the previous and following nodes.

The getKth function helps us find the K-th node efficiently, allowing us to determine whether we have a group of the desired size to reverse.

Conclusion

In this blog post, we’ve addressed the Reverse Nodes In K Group problem on LeetCode.

We provided a detailed Python solution, discussed the problem overview, constraints, and the reasoning behind our approach.

This solution efficiently reverses nodes in groups of size k while handling various edge cases.

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If you have any questions or suggestions, feel free to comment.

Happy coding!

Question Link: Reverse Nodes In K Group on LeetCode

Now that you have a solid understanding

of this problem, give it a try on LeetCode and practice your skills.

Good luck!