The Problem:

In this problem, we are given a linked list and our is to return the head of a modified version of this list, it by merging nodes between two consecutive 0’s into a single node with the sum of their values. The modified list should not contain any 0’s as well.

Before we begin trying to figure out a solution for this problem, let’s first revisit an important concepts Linked List and Recursion.

Understanding Recursion:

Recursion is essentially a function calling itself. It is frequently used to solve problems that can be divided into smaller, similar sub-problems and can be very useful in dynamic programming. To understand recursion, it can be helpful to break it down into 3 parts:

1. Base Condition: This condition prevents the recursive function from calling itself infinitely.
2. Recursive Call: This is responsible for calling the recursive function itself.
3. Small Calculation: This step, which can come before or after the Recursive Call, is responsible for performing some calculation that we need for the recursive function or for returning some result after a recursive call has been completed.

A recursive tree is a visual representation of the recursive calls made by a function. It can be very helpful in understanding how recursion works and in analyzing the time and space complexity of recursive algorithms.

Understanding Linked List:

A linked list is a linear data structure that consists of a sequence of nodes, each containing data and a reference to the next node in the list. It is often used as an alternative to arrays, as it can be more efficient in certain operations such as insertion and deletion.

To understand linked lists, it can be helpful to break it down into 3 parts:

1. Nodes: Each node in a linked list contains some data and a reference to the next node in the list.
2. Head: The head of a linked list is the first node in the list. It is used as the starting point for many operations on the list.
3. Traversal: Traversing a linked list involves starting at the head and following the references from one node to the next until the desired node is reached or the end of the list is reached.

Linked lists can be very useful in solving problems that involve dynamic data structures, as they allow for efficient insertion and deletion of elements. They can also be used in combination with other data structures, such as stacks and queues, to solve more complex problems.

Intuition & Algorithm:

With an understanding of how recursion and linked lists work, let’s try to break down the problem. We don’t need to create our own linked list data structure because it is already provided. So we will begin by figuring out the base condition, which in this case would be when the recursion reaches the end of the linked list (head == None). This means that we have reached the end of the linked list and we can return the start node.

The flag variable is used to keep track of whether we are currently merging nodes or not. When the flag is False, it means that we are not currently merging nodes, and we are simply traversing the linked list until we find a node with value 0. When we find a node with value 0, we set the flag to True to indicate that we have started merging nodes.

If the flag is True, we initialize a sumVal variable to 0 and a startNode variable to head. We then enter a while loop where we keep adding the value of head to sumVal until either flag becomes False or head.next becomes None. The flag becomes False when we encounter another node with value 0, indicating that we have finished merging nodes and we need to update the value of startNode to sumVal and set its next node to head (as by the time we break out of the while loop, the head would point to the next 0).

If head.val is 0, we set flag to True. We then call the helper function recursively on head.next with the updated flag and start node.

After calling the helper function on head, we call another helper function called remove0s on head. This function removes all nodes with a value of 0 from the linked list. It does this by recursively calling itself on head.next until it reaches the end of the linked list (head == None). If head.val is 0, it returns head.next, effectively removing head from the linked list. Otherwise, it returns head.

This ensures that we will always be able to replace nodes between two 0 nodes with their sum and later get rid of the 0 nodes. Let's visualize the algorithm below.

Summary of Solution

Use recursion and a flag variable to merge nodes in a linked list. If flag is True, keep adding the values of nodes until a node with value 0 is encountered. Update the value of the start node to the sum of values and set its next node to head (the next 0 node). Call the helper function recursively on head.next with updated flag and start node. After calling the helper function on head, call another function on head to remove all nodes with value 0 from the linked list.

Code

This can be translated to code in the following way:

# 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]:
        def helper(head, flag,start):
            if head == None:
                return start
            if flag == True:
                sumVal = 0
                startNode = head
                while flag == True and head.next:
                    sumVal += head.val
                    if head.val == 0:
                        flag == False
                        break
                    head = head.next
                startNode.val = sumVal
                startNode.next = head
            if head.val == 0:
                flag = True
            return helper(head.next, flag,start)
            
        head = helper(head, False,head)
        def remove0s(head):
            if head == None:
                return head
            head.next = remove0s(head.next)
            return head.next if head.val == 0 else head
        
        return remove0s(head)

• Time Complexity O(n)
• Space Complexity O(n)

The time complexity of this solution is O(n) This is because the function needs to traverse the entire linked list once to merge the nodes and once again to remove the nodes with value 0. Time taken for this operation would be proportional to the size of the linked list.

Space complexity is O(n) because of recursion call stack.

This approach is simple and efficient, making it a good solution to this problem.

I hope you enjoyed this post. Stay tuned for daily LeetCode blog posts!