Problem Description

In this problem, we are provided with the head node of a singly linked list, which is a data structure where each node contains a value and a reference to the next node. Our task is to remove all consecutive sequences of nodes from this linked list that, when their values are added together, sum up to 0.

The challenge lies in the possibility of such sequences being spread across the linked list and not necessarily being adjacent immediately. Furthermore, after removing a sequence that sums to 0, new sequences can form, also leading to a sum of 0, which should be removed as well. This process must be repeated until there are no more sequences with a sum of 0 remaining.

For example, if we have a linked list 1 -> 2 -> -3 -> 3 -> 1, we can see that the nodes 2 -> -3 -> 3 sum up to 2 + (-3) + 3 = 0. If we remove these nodes, the linked list becomes 1 -> 1. The requirement is to keep removing such sequences until no such sequence exists, and in this case, we would finally end up with an empty linked list as 1 -> 1 also sums up to 0.

Intuition

To solve this problem, we use a two-pass approach leveraging the concept of a prefix sum and a hash table. The prefix sum for a node in the linked list is the sum of all node values from the head up to and including that node. If at any point, we find two nodes with the same prefix sum, the nodes between them sum to 0, and thus that sequence can be removed.

For example, suppose that for a linked list 1 -> 2 -> -3 -> 3 -> 1, the prefix sum at the first 1 is 1, then 3 at 2, back to 0 after -3, up to 3 at the second 3, and finally 4 at the last 1. The appearance of 0 as a prefix sum implies that the subsequence from the beginning up to that point sums to 0.

Here's our approach step by step:

1. We create a dummy node that we'll use as a new head, which helps us deal with edge cases where the beginning of the linked list might sum to 0 and be removed.

2. We iterate through the list, calculating the prefix sum for each node. We keep a track of the last seen node for each prefix sum in a hash table. If the same prefix sum is encountered again, we update it in the hash map with the most recent node. This is because any node sequences between the two nodes with the same sum can be removed, so we only care about the latest position for that prefix sum.

3. After populating the hash table, we make a second pass through the list, again calculating the prefix sum. Using our hash table, we can now update the next pointer of the current node to skip over any nodes that are part of a zero-sum sequence by setting it to the next of the last node remembered for the current prefix sum.

With this approach, we effectively go over the list, find all sequences summing to 0, and remove them, then return the list with all those sequences gone.

Learn more about Linked List patterns.

Solution Approach

The provided solution makes use of the prefix sum concept and a hash table to efficiently find and remove zero-sum consecutive sequences in the linked list. Here's how the implementation unfolds:

1. We initialize a dummy node that acts as a pre-head, ensuring we can handle cases where the head itself might be part of a zero-sum sequence. This dummy node points to the original head of the list.

dummy = ListNode(next=head)

2. We declare a hash table last to record the last node for each unique prefix sum observed. The hash table is indexed by the prefix sum and contains the corresponding node as its value.

3. Starting from the dummy node (pre-head), we iterate through the linked list to calculate the prefix sum s for each node. As we compute the prefix sum, we update the last hash table with the current node. If the same sum occurs again later, it overwrites the previous node, since we only need the latest one.

s, cur = 0, dummy
while cur:
    s += cur.val
    last[s] = cur
    cur = cur.next

4. After populating the last table, we iterate the list again, starting from the dummy node, to update the next pointers. For each node, as we calculate the prefix sum s, we find the node corresponding to this sum in the last table, and we set the current node’s next pointer to last[s].next. This effectively skips over and removes any nodes part of a zero-sum sequence found between the two nodes with equal prefix sums.

s, cur = 0, dummy
while cur:
    s += cur.val
    cur.next = last[s].next
    cur = cur.next

5. Finally, we return dummy.next — the head of the modified list, which no longer contains any sequence of nodes that sum up to 0.

The two primary components used in the solution are:

• Prefix Sum: This technique is critical to discover sequences that total to 0. By keeping track of the cumulative sum at each node, we can swiftly identify regions of the list that cancel each other out.
• Hash Table: By storing the last occurrence of a node for a given prefix sum, we have the ability to quickly jump over sequences that sum to 0. This is because if a prefix sum repeats, the sum of the nodes between those repetitions is necessarily 0.

Together, these structures allow the algorithm to achieve its goal with a linear time complexity relative to the number of nodes in the list, since each node is processed directly without the need for nested loops or recursion.

Example Walkthrough

Let's consider a simple linked list to illustrate the solution approach: 3 -> 4 -> -7 -> 5 -> -6 -> 6. The goal is to remove sublists that sum to 0.

1. Initialize a dummy node and a last hash table

   We create a dummy node and initialize our last hash table to store the last node associated with each unique prefix sum. Initially, the dummy node points to the head of our list (3).

2. First pass: Compute prefix sums and populate the last table

   Starting from the dummy node, the prefix sum (s) and last table will be updated as follows:

   s: 0 -> dummy node (There's always a dummy node associated with prefix sum 0)
   s: 3 -> 3 (First node with value 3)
   s: 7 -> 4 (Node with value 4)
   s: 0 -> -7 (Node with value -7, this gives us a prefix sum of 0, meaning everything from the dummy to this node sums to 0)
   s: 5 -> 5 (We continue the process from the node with value 5)
   s: -1 -> -6
   s: 5 -> 6 (The last node, summing up to 5)

   Notice how the prefix sum returned to 0 when we included the -7 node, implying the sublist 3 -> 4 -> -7 sums to 0 and should be removed.

3. Second pass: Update the next pointers using the last table

   We iterate through the list again, using the last table to update next pointers. We perform the following updates, recalculating the prefix sum s:

   s: 0, dummy.next -> last[0].next (skipping to node with value 5)
   s: 5, 5.next -> last[5].next (skipping to node with value 6)

   After the second pass, the list becomes 5 -> 6, since the first part 3 -> 4 -> -7 has been skipped. We then continue and find that 5 -> -6 -> 6 also sum to 0 and should be removed.

4. Final list

   After all updates, the prefix sums that led to a non-zero result have been removed, and we are left with an empty list, as the entire list was a combination of zero-sum sublists. The dummy.next now points to None.

By using the prefix sum and a hash table to keep track of the last occurrence of each prefix sum, we have efficiently removed consecutive sequences that sum to 0. The resulting list would be returned as the modified list without these zero-sum sequences.

Solution Implementation

Time and Space Complexity

The provided code has a time complexity of O(n) where n is the length of the linked list. This is because the code consists of two separate while loops that each iterate through the list once. The first while loop constructs a dictionary (last) mapping the cumulative sums of the nodes up to that point to the corresponding node. The second while loop uses the dictionary to skip over nodes that are part of a zero-sum sublist by setting the next pointer of previous non-zero-sum nodes to the next node in the last occurrence of that sum.

The space complexity of the code is O(n) as well. The primary factor contributing to the space complexity is the dictionary (last) which stores a value for each unique cumulative sum encountered while iterating through the list. In the worst-case scenario, there could be as many unique sums as there are nodes in the list (when no sublists sum to zero), which would require storing each node in the dictionary, hence O(n) space is needed.

Learn more about how to find time and space complexity quickly using problem constraints.