<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>

class Solution {
public:
    void solve(vector<int>&v,int k)
    {
        int n = v.size();
        for(int i = 0; i < n; i += k)
        {
            if(i + k <= n)  // Only reverse if a full group of size k can be formed
            {
                reverse(v.begin() + i, v.begin() + i + k);
            }
        }
    }

    ListNode* reverseKGroup(ListNode* head, int k) 
    {
        if (k <= 1 || head == nullptr) return head; // If k is 1 or head is NULL, no need to reverse

        vector<int>v;
        ListNode* curr = head;
        while(curr != nullptr)
        {
            v.push_back(curr->val);
            curr = curr->next;
        }

        solve(v, k); // Reverse groups of k

        // Reconstruct the linked list from the modified vector
        curr = head;
        int i = 0;
        while(curr != nullptr)
        {
            curr->val = v[i++];
            curr = curr->next;
        }

        return head;
    }
};

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