# 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 class ReverseEveryKElements {

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

ListNode prev = null;

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

if (count == k) {
if (prev == null) {
} else {
}

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

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;
}``````
1 Like

I suppose the course’s staff has already fixed the problem’s statement:

If, in the end, you are left with a sub-list with less than ‘k’ elements, reverse it too.

Actually, the ask to not reverse last set of nodes less than k, is a Hard question in leetcode (#25).

We can just put a check to have minimum k nodes to achieve same. Posting the code in case anyone needs it.

``````public ListNode reverseKGroup(ListNode head, int k) {

if(k<=1)

ListNode previousNode = null;
ListNode nextNode = null;

ListNode lastNodeOfFirstPart = null;
ListNode startNodeOfSecondPart = null;

//result must be reurned as first kth Node, for 1->2>3->4  it will be 2 if K==2 OR it will be 3 if k==3;

while(true){

lastNodeOfFirstPart = previousNode;
startNodeOfSecondPart = currentNode;//it will be 1,3,5, for k==2

//check the number of nodes available to start with, if less than k break
int nodecounter=0;
ListNode tempCurrent = currentNode;
while(tempCurrent!=null && nodecounter <=k){
nodecounter++;
tempCurrent = tempCurrent.next;
}

if(nodecounter<k)
break;   //breaking if less than k nodes.

for(int i=1; currentNode!=null && i<=k;i++){
nextNode = currentNode.next;
currentNode.next = previousNode;
previousNode = currentNode;
currentNode = nextNode;
}

if(lastNodeOfFirstPart!=null)
lastNodeOfFirstPart.next = previousNode;
else
head = previousNode;//for first iteration this will be the last node of firt K group.

startNodeOfSecondPart.next = currentNode; //this is tricky??
//above is needed to have link  between firs reversed pard and 2nd start part ->1->3

previousNode = startNodeOfSecondPart;//This is the 1->2  1 here.

if(currentNode==null){
//System.out.println("currentNode is NULL-->");
break;
}
}