Solution Explanation: Remove Zero Sum Consecutive Nodes from Linked List

This problem involves removing consecutive sequences of nodes in a linked list that sum to zero. The optimal approach uses a hash table (or dictionary) to efficiently track prefix sums and identify zero-sum subsequences.

Approach:

1. Prefix Sum: We iterate through the linked list, calculating the prefix sum at each node. The prefix sum at a node is the sum of all node values from the head of the list up to that node.

2. Hash Table (Dictionary): We use a hash table to store the prefix sum and the corresponding node. The key is the prefix sum, and the value is a pointer to the node where that prefix sum occurs. If a prefix sum is encountered multiple times, we only keep the latest occurrence (overwriting the earlier one). This is because we're only interested in the last node that results in a particular prefix sum.

3. Zero-Sum Detection: If we encounter a prefix sum s that already exists in the hash table, it means the sum of the nodes between the node associated with the previous s and the current node is zero (because the prefix sums are equal).

4. Node Removal: To remove the zero-sum subsequence, we directly modify the next pointer of the node associated with the previous s to point to the current node's next node, effectively skipping the zero-sum subsequence.

Code Explanation (Python):

# 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]:
        dummy = ListNode(next=head)  # Dummy node simplifies handling the head
        prefix_sums = {0: dummy} # Initialize with prefix sum 0 at dummy node
        current_sum = 0
        curr = dummy
 
        while curr:
            current_sum += curr.val
            if current_sum in prefix_sums:
                #Zero sum subsequence found
                prev_node = prefix_sums[current_sum]
                prev_node.next = curr.next #Skip the subsequence
                prefix_sums = {0:dummy} #Reset the prefix sum dictionary
                current_sum=0
                curr = dummy
            else:
                prefix_sums[current_sum] = curr
                curr = curr.next
        return dummy.next
 

Time Complexity: O(N), where N is the number of nodes in the linked list. We iterate through the list twice in the worst case.

Space Complexity: O(N) in the worst case to store the prefix sums in the hash table. In the best case (no zero-sum subsequences), it's O(N).

Other Languages:

The solutions provided in Java, C++, Go, Typescript and Rust follow the same logic, only adapting the syntax and data structures specific to each language. The core idea of using prefix sums and a hash table remains consistent. The primary difference is how hash maps are implemented in each language (e.g., HashMap in Java, unordered_map in C++, map in Go).

Example Walkthrough:

Let's trace the Python code with head = [1, 2, -3, 3, 1]:

1. prefix_sums = {0: dummy} (dummy node)
2. Iterate:
   • 1: current_sum = 1, prefix_sums = {0: dummy, 1: node1}
   • 3: current_sum = 3, prefix_sums = {0: dummy, 1: node1, 3: node2}
   • 0: current_sum = 0, 0 is in prefix_sums, so we remove [1, 2, -3]. prefix_sums is reset, and the iteration restarts.
   • 3: current_sum = 3, prefix_sums = {0: dummy, 3: node3}
   • 4: current_sum = 4, prefix_sums = {0: dummy, 3: node3, 4: node4}
3. The final list becomes [3, 1].

This approach efficiently solves the problem by avoiding repeated scans of the list and directly modifying pointers to remove zero-sum subsequences.