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Incorrect solution for Reverse every K-element Sub-list

The solution given by the author seems to give us the wrong solution. The reason is when a sublist has less then k nodes it still reverses it.
Eg: Given n = 1,2,3,4,5,6,7,8 & k = 3
first we would reverse the first k nodes
n = 3-2-1-4-5-6-7-8
Then we would reverse the nest k nodes
n = 3-2-1-6-5-4-7-8
This should the actual solution, but the answer given by the admin seems to reverse the next 2 nodes as well.
n = 3-2-1-6-5-4-8-7

Instead we should first iterate over the list k nodes and see if there are enough nodes to reverse and then we reverse the first set of nodes. Repeat the process till there are k possible nodes to reverse, if not then add rest of the remaining nodes without reversing them.

Here’s the right code :slight_smile:
class ReverseEveryKElements {

  public static ListNode reverse(ListNode head, int k) {
    if (k < 2 || head == null) {
      return head;
    }

    ListNode curr = head;
    ListNode itrHead = head;
    ListNode prev = null;

    while (itrHead != null) {
      int count = 0;
      for (int i = 0; i < k && itrHead != null; i++) {
        itrHead = itrHead.next;
        count++;
      }

      if (count == k) {
        ListNode rev_head = reverseSublist(curr, k);
        if (prev == null) {
          head = rev_head;
        } else {
          prev.next = rev_head;
        }

        prev = curr;
        curr = itrHead;
      } else {
        if (prev != null) {
          prev.next = curr;
        }
      }
    }
    return head;
  }

  public static ListNode reverseSublist(ListNode node, int k) {
    ListNode curr = node;
    ListNode prev = null;
    ListNode next = null;

    for (int i = 0; i < k && curr != null; i++) {
      next = curr.next;
      curr.next = prev;
      prev = curr;
      curr = next;
    }

    return prev;
  }
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