#1171

Medium

auto_awesomeLinked-list pointer manipulation

LeetCode Problem Workspace

Remove Zero Sum Consecutive Nodes from Linked List

Remove zero-sum consecutive nodes from a linked list by iterating through it, repeatedly deleting sequences that sum to zero.

hash table linked list

Problem Statement

Given the head of a linked list, repeatedly remove consecutive sequences of nodes that sum to zero until no such sequences remain. After all deletions, return the head of the resulting list. This can be done with a straightforward traversal, and there are multiple valid outputs based on the sequences removed.

For example, with the input [1,2,-3,3,1], the zero-sum sequence [1,2,-3] can be deleted, leaving [3,1]. The output can vary, with an alternative solution being [1,2,1]. The problem ensures that any list with consecutive zero-sum sequences is reduced to its final form, returning the head of the modified list.

Examples

Example 1

Input: head = [1,2,-3,3,1]

Output: [3,1] Note: The answer [1,2,1] would also be accepted.

Example details omitted.

Example 2

Input: head = [1,2,3,-3,4]

Output: [1,2,4]

Example details omitted.

Example 3

Input: head = [1,2,3,-3,-2]

Output: [1]

Example details omitted.

Constraints

The given linked list will contain between 1 and 1000 nodes.
Each node in the linked list has -1000 <= node.val <= 1000.

Solution Approach

Using a Hash Map to Track Sums

We traverse the linked list while maintaining a running sum and storing the sum at each node in a hash map. If a sum repeats, it means there's a zero-sum sequence between those two nodes, which can be removed by adjusting pointers.

Pointer Manipulation to Remove Nodes

By manipulating the linked list pointers, we can skip over nodes forming the zero-sum sequence. This is done by finding the previous occurrence of the sum in the hash map and updating the pointer to bypass the zero-sum nodes.

Final Pass for Clean List

After detecting all zero-sum sequences, we perform a final pass through the list to ensure that all consecutive sequences are removed and return the head of the cleaned-up list.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(n)

The time complexity is O(n) due to the single pass through the list, where n is the number of nodes. The space complexity is also O(n) because we store the sum at each node in a hash map, which grows with the size of the list.

What Interviewers Usually Probe

The candidate should demonstrate understanding of linked-list traversal and pointer manipulation.
Look for the ability to optimize using a hash map to track sums efficiently.
Evaluate how well the candidate handles edge cases such as lists with no zero-sum sequences or lists that are completely zero-sum.

Common Pitfalls or Variants

Common pitfalls

Failing to properly update pointers when removing zero-sum sequences, leading to lost nodes.
Overlooking edge cases like empty lists or lists without any zero-sum sequences.
Misusing the hash map to track sums, which can cause incorrect removal of nodes.

Follow-up variants

Using a different approach for removing zero-sum sequences, such as a stack-based solution.
Handling cases where the linked list contains only one node or very small lists.
Returning multiple possible outputs based on how the zero-sum sequences are removed.

FAQ

What is the approach for solving the 'Remove Zero Sum Consecutive Nodes from Linked List' problem?

The solution involves iterating through the linked list while tracking cumulative sums using a hash map, adjusting pointers when zero-sum sequences are detected.

Can the problem have multiple valid outputs?

Yes, depending on which zero-sum sequences are removed, multiple valid outputs are possible, as long as all zero-sum nodes are excluded from the result.

What is the time complexity of the solution?

The time complexity is O(n), where n is the number of nodes in the linked list, as the list is traversed once.

How does the hash map help in solving this problem?

The hash map stores cumulative sums as we traverse the list. When the same sum appears again, it indicates a zero-sum sequence, which can be removed efficiently.

What are some common pitfalls when solving this problem?

Common mistakes include failing to update pointers correctly, not handling edge cases like empty lists, or misusing the hash map to track sums.

terminal

Solution

Solution 1: Prefix Sum + Hash Table

If two prefix sums of the linked list are equal, it means that the sum of the continuous node sequence between the two prefix sums is $0$, so we can remove this part of the continuous nodes.

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# 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)
        last = {}
        s, cur = 0, dummy
        while cur:
            s += cur.val
            last[s] = cur
            cur = cur.next
        s, cur = 0, dummy
        while cur:
            s += cur.val
            cur.next = last[s].next
            cur = cur.next
        return dummy.next

Continue Practicing

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