LeetCode #2181 — MEDIUM

Merge Nodes in Between Zeros

Move from brute-force thinking to an efficient approach using linked list strategy.

The Problem

Problem Statement

You are given the head of a linked list, which contains a series of integers separated by 0's. The beginning and end of the linked list will have Node.val == 0.

For every two consecutive 0's, merge all the nodes lying in between them into a single node whose value is the sum of all the merged nodes. The modified list should not contain any 0's.

Return the head of the modified linked list.

Example 1:

Input: head = [0,3,1,0,4,5,2,0]
Output: [4,11]
Explanation: 
The above figure represents the given linked list. The modified list contains
- The sum of the nodes marked in green: 3 + 1 = 4.
- The sum of the nodes marked in red: 4 + 5 + 2 = 11.

Example 2:

Input: head = [0,1,0,3,0,2,2,0]
Output: [1,3,4]
Explanation: 
The above figure represents the given linked list. The modified list contains
- The sum of the nodes marked in green: 1 = 1.
- The sum of the nodes marked in red: 3 = 3.
- The sum of the nodes marked in yellow: 2 + 2 = 4.

Constraints:

The number of nodes in the list is in the range [3, 2 * 10⁵].
0 <= Node.val <= 1000
There are no two consecutive nodes with Node.val == 0.
The beginning and end of the linked list have Node.val == 0.

Step 01

Brute Force Baseline

Problem summary: You are given the head of a linked list, which contains a series of integers separated by 0's. The beginning and end of the linked list will have Node.val == 0. For every two consecutive 0's, merge all the nodes lying in between them into a single node whose value is the sum of all the merged nodes. The modified list should not contain any 0's. Return the head of the modified linked list.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Linked List

Example 1

[0,3,1,0,4,5,2,0]

Example 2

[0,1,0,3,0,2,2,0]

Core Insight

What unlocks the optimal approach

How can you use two pointers to modify the original list into the new list?
Have a pointer traverse the entire linked list, while another pointer looks at a node that is currently being modified.
Keep on summing the values of the nodes between the traversal pointer and the modifying pointer until the former comes across a ‘0’. In that case, the modifying pointer is incremented to modify the next node.
Do not forget to have the next pointer of the final node of the modified list point to null.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2181: Merge Nodes in Between Zeros
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeNodes(ListNode head) {
        ListNode dummy = new ListNode();
        int s = 0;
        ListNode tail = dummy;
        for (ListNode cur = head.next; cur != null; cur = cur.next) {
            if (cur.val != 0) {
                s += cur.val;
            } else {
                tail.next = new ListNode(s);
                tail = tail.next;
                s = 0;
            }
        }
        return dummy.next;
    }
}

// Accepted solution for LeetCode #2181: Merge Nodes in Between Zeros
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeNodes(head *ListNode) *ListNode {
	dummy := &ListNode{}
	tail := dummy
	s := 0
	for cur := head.Next; cur != nil; cur = cur.Next {
		if cur.Val != 0 {
			s += cur.Val
		} else {
			tail.Next = &ListNode{Val: s}
			tail = tail.Next
			s = 0
		}
	}
	return dummy.Next
}

# Accepted solution for LeetCode #2181: Merge Nodes in Between Zeros
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy = tail = ListNode()
        s = 0
        cur = head.next
        while cur:
            if cur.val:
                s += cur.val
            else:
                tail.next = ListNode(s)
                tail = tail.next
                s = 0
            cur = cur.next
        return dummy.next

// Accepted solution for LeetCode #2181: Merge Nodes in Between Zeros
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn merge_nodes(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut dummy = Box::new(ListNode::new(0));
        let mut tail = &mut dummy;
        let mut s = 0;
        let mut cur = head.unwrap().next;

        while let Some(mut node) = cur {
            if node.val != 0 {
                s += node.val;
            } else {
                tail.next = Some(Box::new(ListNode::new(s)));
                tail = tail.next.as_mut().unwrap();
                s = 0;
            }
            cur = node.next.take();
        }

        dummy.next
    }
}

// Accepted solution for LeetCode #2181: Merge Nodes in Between Zeros
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function mergeNodes(head: ListNode | null): ListNode | null {
    const dummy = new ListNode();
    let tail = dummy;
    let s = 0;
    for (let cur = head.next; cur; cur = cur.next) {
        if (cur.val) {
            s += cur.val;
        } else {
            tail.next = new ListNode(s);
            tail = tail.next;
            s = 0;
        }
    }
    return dummy.next;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.