<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]:
        """
        If we see with little focus we will come to know that
        we are removing the nodes with sum value 0.
        But here the question is how we will come to know that
        sum of subset nodes is 0. For this we can use prefix sum.
        If we are having sum which has already appeared once before
        it means that we are having sum of nodes between them = 0.
        """
        dummy = ListNode(0, head)
        hashmap = {0: dummy}
        prefix_sum = 0

        while head:
            prefix_sum += head.val
            hashmap[prefix_sum] = head
            head = head.next

        head = dummy
        prefix_sum = 0
        while head:
            prefix_sum += head.val
            head.next = hashmap[prefix_sum].next
            head = head.next

        return dummy.next
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