<p>You are given the <code>head</code> of a linked list.</p>

<p>Remove every node which has a node with a greater value anywhere to the right side of it.</p>

<p>Return <em>the </em><code>head</code><em> of the modified linked list.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/10/02/drawio.png" style="width: 631px; height: 51px;" />
<pre>
<strong>Input:</strong> head = [5,2,13,3,8]
<strong>Output:</strong> [13,8]
<strong>Explanation:</strong> The nodes that should be removed are 5, 2 and 3.
- Node 13 is to the right of node 5.
- Node 13 is to the right of node 2.
- Node 8 is to the right of node 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [1,1,1,1]
<strong>Output:</strong> [1,1,1,1]
<strong>Explanation:</strong> Every node has value 1, so no nodes are removed.
</pre>

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

<ul>
	<li>The number of the nodes in the given list is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    # def removeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
    #     stack = []
    #     temp = head
    #     prev = dummy = ListNode(0)
    #     prev_prev = None

    #     while temp:
    #         while stack and stack[-1].val < temp.val:
    #             stack.pop()
    #         stack.append(temp)
    #         temp = temp.next

    #     nxt = None
    #     while stack:
    #         popped = stack.pop()
    #         popped.next = nxt
    #         nxt = popped

    #     return popped
                

    def removeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
        new_head = temp = self.reverse(head)
        while temp.next:
            if temp.next.val >= temp.val:
                temp = temp.next
            else:
                temp.next = temp.next.next
        return self.reverse(new_head)

    def reverse(self, head):
        nxt = None
        temp = head
        while temp:
            next_node = temp.next
            temp.next = nxt
            nxt = temp
            temp = next_node
        return nxt



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