LeetCode - Reverse Nodes in k-Group

Problem statement

Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list's nodes, only nodes themselves may be changed.

Problem statement taken from: https://leetcode.com/problems/reverse-nodes-in-k-group

Example 1:

Container

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

Example 2:

Container

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

Constraints:

- The number of nodes in the list is n.
- 1 <= k <= n <= 5000
- <= Node.val <= 1000

Follow-up: Can you solve the problem in O(1) extra memory space?

Explanation

Reverse a linked list

We can use the standard Reverse linked list code with a slight modification. We pass the count k to the method, which reverses the sublist of size k. We should keep the track of the next node and the previous node. These are required to point the pointers of the current sublist correctly to our previous sublist.

A C++ snippet of this approach is as follows:

ListNode* reverse(ListNode* head, int k) {
    if (!head)
        return NULL;

    ListNode* current = head;
    ListNode* next = NULL;
    ListNode* prev = NULL;
    int count = 0;

    while (current != NULL && count < k) {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;
        count++;
    }

    if (next != NULL)
        head->next = reverse(next, k);

    return prev;
}

The time complexity of this approach is O(n). The space complexity is O(n / k). For a linked list of size n, we make n/k or n/k + 1 calls during recursion.

Optimized solution: Iterative

We can optimize the space by using the above approach without recursion. We keep track of the previous, current, and next nodes while reversing the linked list in a set of size k. Once the sublist of size k is reversed we update the previous, current, and next node correctly. We repeat this approach till the list is traversed or the last sublist is less than size k.

Let's check the algorithm

Algorithm

- if !head || k == 1
  - return head

- set ListNode *temp = new ListNode(1)
    temp->next = head

- set ListNode *prev, *current, *next = temp
  set count = 0
  initialize index and i variables

// count the size of the list
- loop while current
  - current = current->next
  - count++
- while end

- loop while next
  - set current = prev->next
  - set next = current->next

  // if the last sublist is less than size k
  // we keep the list as it is.
  // Hence setting index = 0.
  - index = count > k ? k : 0

  - loop for i = 1; i < index; i++
    - set current->next = next->next
      next->next = prev->next
      prev->next = next
      next = current->next
  - for end

  - set prev = current

  - update count = count - k
- for end

- return temp->next

The time complexity of the above approach is O(n). The space complexity is O(1).

Let's check our algorithm in C++, Golang, and JavaScript.

C++ solution

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        if(!head || k == 1) {
            return head;
        }

        ListNode *temp = new ListNode(1);
        temp->next = head;

        ListNode *prev = temp, *current = temp, *next = temp;

        int count = 0, index, i;

        while(current) {
            current = current->next;
            count++;
        }

        while(next) {
            current = prev->next;
            next = current->next;

            index = count > k ? k : 0;

            for(i = 1; i < index; i++) {
                current->next = next->next;
                next->next = prev->next;
                prev->next = next;
                next = current->next;
            }

            prev = current;

            count -= k;
        }

        return temp->next;
    }
};

Golang solution

func reverseKGroup(head *ListNode, k int) *ListNode {
    if head == nil || k == 1 {
        return head
    }

    temp := &ListNode{1, nil}
    temp.Next = head

    prev, current, next := temp, temp, temp
    count, index, i := 0, 0, 0

    for current != nil {
        current = current.Next
        count++
    }

    for next != nil {
        current = prev.Next
        next = current.Next

        if count > k {
            index = k
        } else {
            index = 0
        }

        for i = 1; i < index; i++ {
            current.Next = next.Next
            next.Next = prev.Next
            prev.Next = next
            next = current.Next
        }

        prev = current
        count -= k
    }

    return temp.Next
}

JavaScript solution

var reverseKGroup = function(head, k) {
    if(!head || k == 1) {
        return head;
    }

    let temp = new ListNode(1, null);
    temp.next = head;

    let prev = temp, current = temp, next = temp;
    let count = 0, index, i;

    while(current) {
        current = current.next;
        count++;
    }

    while(next) {
        current = prev.next;
        next = current.next;

        index = count > k ? k : 0;

        for(i = 1; i < index; i++) {
            current.next = next.next;
            next.next = prev.next;
            prev.next = next;
            next = current.next;
        }

        prev = current;

        count -= k;
    }

    return temp.next;
};

Let's dry-run our algorithm to see how the solution works.

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

Step 1: if !head || k == 1
          head -> [1, 2, 3, 4, 5] || 3 == 1
          false

Step 2: temp = new ListNode(1)
             -> [1]

        temp->next = head

        temp -> [1, 1, 2, 3, 4, 5]
        head -> [1, 2, 3, 4, 5]

Step 3: ListNode *prev = temp, *current = temp, *next = temp
        count = 0
        index, i

Step 4: loop while current
          current = current->next
          count++

        This will count the size of linked list.
        count = 5

Step 5: loop while next
          current = prev->next
                  = [1, 2, 3, 4, 5]

          next = current->next
               = [2, 3, 4, 5]

          index = count > k ? k : 0
                = 5 > 2 ? 2 : 0
                = 2

          loop for i = 1; i < 2
            1 < 2
            true

            current->next = next->next
                          = [3, 4, 5]

            next->next = prev->next
                       = [1, 3, 4, 5]

            prev->next = next
                       = [2, 1, 3, 4, 5]

            next = current->next
                 = [3, 4, 5]

            i++
            i = 2

          loop for i < 2
            2 < 2
            false

          prev = current
               = [1, 3, 4, 5]

          count = count - k
                = 5 - 2
                = 3

Step 6: loop while next
          next = [3, 4, 5]

          current = prev->next
                  = [3, 4, 5]

          next = current->next
               = [4, 5]

          index = count > k ? k : 0
                = 3 > 2 ? 2 : 0
                = 2

          loop for i = 1; i < 2
            1 < 2
            true

            current->next = next->next
                          = [3, 5]

            next->next = prev->next
                       = [1, 4, 5]

            prev->next = next
                       = [4, 3, 5]

            next = current->next
                 = [5]

            i++
            i = 2

          loop for i < 2
            2 < 2
            false

          prev = current
               = [5]

          count = count - k
                = 3 - 2
                = 1

          temp = [2, 1, 4, 3, 5]

Step 7: loop while next
          next = [5]

          current = prev->next
                  = [5]

          next = current->next
               = nil

          index = count > k ? k : 0
                = 1 > 2 ? 2 : 0
                = 0

          loop for i = 1; i < 0
            1 < 0
            false

          prev = current
               = [5]

          count = count - k
                = 1 - 2
                = -1

Step 8: loop while next
          next = nil
          false

Step 9: return temp->next
        temp = [1, 2, 1, 4, 3, 5]
        temp->next = [2, 1, 4, 3, 5]

We return the answer as [2, 1, 4, 3, 5].

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