<p>Given the <code>head</code> of a linked list, reverse the nodes of the list <code>k</code> at a time, and return <em>the modified list</em>.</p>

<p><code>k</code> 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 <code>k</code> then left-out nodes, in the end, should remain as it is.</p>

<p>You may not alter the values in the list&#39;s nodes, only nodes themselves may be changed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], k = 2
<strong>Output:</strong> [2,1,4,3,5]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], k = 3
<strong>Output:</strong> [3,2,1,4,5]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is <code>n</code>.</li>
	<li><code>1 &lt;= k &lt;= n &lt;= 5000</code></li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow-up:</strong> Can you solve the problem in <code>O(1)</code> extra memory space?</p>

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* getkthNode(ListNode* temp, int k)
    {
        k--;
        
        while(temp != NULL && k > 0)
        {
            k--;
            temp = temp->next;
        }
        
        return temp;
    }
    
    ListNode* reverseKGroup(ListNode* head, int k) {
        
        if(head == nullptr)
            return head;
        
        ListNode* curr = head;
        ListNode* newHead = NULL;
        ListNode* tail = NULL;
        
        while(curr != NULL)
        {
            //keep reversing till k nodes
            ListNode* prev = NULL;
            ListNode* groupStart = curr;
            int count = 0;
            
            if(getkthNode(groupStart,k) == NULL)
            {
                if(tail)
                tail->next = groupStart;
                break;
            }
            
            while(curr != NULL && count < k)
            {
                ListNode* temp = curr -> next;
                curr -> next = prev;
                prev = curr;
                curr = temp;
                count++;
            }

            if(newHead == NULL)  // in case of first group reverse
            {
                newHead = prev;
            }

            if(tail != NULL)
            {
                tail -> next = prev;
            }

            tail = groupStart;    
        }

        return newHead;
        
    }
};
16500