<p>Given the <code>head</code> of a linked list, we repeatedly delete consecutive sequences of nodes that sum to <code>0</code> until there are no such sequences.</p>

<p>After doing so, return the head of the final linked list.&nbsp; You may return any such answer.</p>

<p>&nbsp;</p>
<p>(Note that in the examples below, all sequences are serializations of <code>ListNode</code> objects.)</p>

<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2,-3,3,1]
<strong>Output:</strong> [3,1]
<strong>Note:</strong> The answer [1,2,1] would also be accepted.
</pre>

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

<pre>
<strong>Input:</strong> head = [1,2,3,-3,4]
<strong>Output:</strong> [1,2,4]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2,3,-3,-2]
<strong>Output:</strong> [1]
</pre>

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

<ul>
	<li>The given linked list will contain between <code>1</code> and <code>1000</code> nodes.</li>
	<li>Each node in the linked list has <code>-1000 &lt;= node.val &lt;= 1000</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 removeZeroSumSublists(self, head: Optional[ListNode]) -> Optional[ListNode]:
    '''
    look for num that will add to 0: 1 -> -1, 2 -> -2
    consecutive: prefix sum
    modified two sum, target num is 0
    '''
    # if not head or not head.next:
    #   return head
    
    dummy = ListNode(0)
    dummy.next = head
    prefix = 0
    prefix_sum = {0: dummy}
    curr = head

    while curr:
      prefix += curr.val
      print(prefix, prefix_sum.keys())
      if prefix in prefix_sum:
        delete = prefix_sum[prefix].next
        print(delete.val)
        temp = prefix + delete.val
        while delete != curr:
          del prefix_sum[temp]
          delete = delete.next
          temp += delete.val
        prefix_sum[prefix].next = curr.next
      else:
        prefix_sum[prefix] = curr
      curr = curr.next
    return dummy.next
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