How do you solve Merge Nodes in Between Zeros on LeetCode?

Merge Nodes in Between Zeros (LeetCode #2181) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The problem requires careful handling of linked list nodes and understanding of how to traverse and modify them.

What is the brute force solution for Merge Nodes in Between Zeros?

The brute force approach for Merge Nodes in Between Zeros is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves traversing the linked list multiple times to sum the values between every pair of zeros. This is simple but inefficient because we repeatedly visit nodes. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Merge Nodes in Between Zeros?

Merge Nodes in Between Zeros uses the following concepts: Linked List, Simulation. The recommended patterns to study are: Two Pointers, Linked List Manipulation.

What are the interview tips for Merge Nodes in Between Zeros?

Always clarify the problem and constraints before jumping into coding. Think about edge cases and how your solution will handle them. Practice explaining your thought process as you code; it helps you think clearly and shows your understanding.

What are common mistakes when solving Merge Nodes in Between Zeros?

Students often forget to reset the sum after creating a new node. Some may not handle the edge cases correctly, such as ensuring no nodes with value 0 are included in the final list.

#2181

Merge Nodes in Between Zeros

Medium

Linked ListSimulationTwo PointersLinked List Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a single pass through the linked list to sum values between zeros, thus improving efficiency. We maintain a running sum and only create new nodes when we encounter a zero.

⚙️

Algorithm

4 steps

1Step 1: Initialize a dummy node to help build the new list and a pointer to traverse the original list.
2Step 2: Use a variable to keep track of the sum of values between zeros.
3Step 3: When a zero is encountered, if the sum is greater than zero, create a new node with the sum and link it to the new list.
4Step 4: Reset the sum and continue until the end of the list.

solution.py20 lines

1class ListNode:
2    def __init__(self, val=0, next=None):
3        self.val = val
4        self.next = next
5
6def mergeNodes(head):
7    current = head.next
8    dummy = ListNode(0)
9    tail = dummy
10    sum_nodes = 0
11    while current:
12        if current.val == 0:
13            if sum_nodes > 0:
14                tail.next = ListNode(sum_nodes)
15                tail = tail.next
16            sum_nodes = 0
17        else:
18            sum_nodes += current.val
19        current = current.next
20    return dummy.next

ℹ

Complexity note: The time complexity is O(n) because we traverse the list only once. The space complexity is O(n) due to the new linked list we create to store the sums.

1The problem requires careful handling of linked list nodes and understanding of how to traverse and modify them.
2Using a dummy node simplifies the process of building a new linked list.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.