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# Definition for singly-linked list.
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class ListNode:
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    def __init__(self, val=0, next=None):
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        self.val = val
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        self.next = next
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class Solution:
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    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
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        # Intialize the result and the current node to the head
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        res = node = head
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        # Initialize the index of the current node to 0
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        i = 0
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        # Initialize the head and tail of the reversed nodes group to None
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        reversedHead, reversedTail = None, None
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        # Initialize the tail of the previous group to None
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        previousTail = None
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        # Iterate through all nodes
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        while node:
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            # When we reach the first node in a group
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            if i % k == 0:
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                # If there is a previous group, connect its tail to the current node
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                # This is the case when we have less than k nodes left
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                if previousTail:
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                    previousTail.next = node
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                # Initialize the head and tail of the reversed nodes group
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                reversedHead = reversedTail = ListNode(node.val)
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            # Continue to reverse subsequent nodes
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            else:
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                reversedHead = ListNode(node.val, reversedHead)
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            # If we are able to reach the last node in a reversed nodes group
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            if i % k == k - 1:
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                # If there is a previous group, connect its tail to the current node
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                # This is the case when we have k nodes and thus, we should reverse this group
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                if previousTail:
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                    previousTail.next = reversedHead
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                # Set the tail of the previous group to the tail of the reversed nodes group
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                previousTail = reversedTail
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            # Set the head of the first reversed nodes group as the result
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            if i == k - 1:
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                res = reversedHead
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            # Continue to the next node
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            i, node = i + 1, node.next
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        return res