Merge pull request #3 from paxtonfitzpatrick/main

solution and notes for problem 2181

Author: jeremymanning

README.md

@@ -8,6 +8,7 @@ Each day (ideally) we'll attempt the daily [leetcode](https://leetcode.com) prob
 |--------------|-------------------------------------------------------------------------------------------------------------|-----------------------------------------------------------------------------------------------------------|------------|
 | July 1, 2024 | [350](https://leetcode.com/problems/intersection-of-two-arrays-ii/)                                         | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/350)                     | Easy       |
 | July 2, 2024 | [1509](https://leetcode.com/problems/minimum-difference-between-largest-and-smallest-value-in-three-moves/) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1509)                    | Medium     |
+| July 3, 2024 | [2181](https://leetcode.com/problems/merge-nodes-in-between-zeros/)                                         | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/2181)                    | Medium     |
 
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problems/2181/paxton.md

@@ -0,0 +1,54 @@
+# [Problem 2181: Merge Nodes in Between Zeros](https://leetcode.com/problems/merge-nodes-in-between-zeros/description)
+
+## Initial thoughts (stream-of-consciousness)
+- start off by iterating through the linked list from the head node
+- whenever we encounter a node with a `.val` of 0, start incrementing a running total of node values beginning with the next node
+- when we reach another 0-value node:
+  - construct a `ListNode` whose `.val` is the running total (and `.next` is `None`, for now)
+  - if this is the first `ListNode` we've created, it's the head node we need to return at the end...
+  - ...otherwise, the last node we made needs to point to it -- so set *that* node's `.next` to this new `ListNode`
+  - store the newly created `ListNode` so that we can update its `.next` the next time we create a new node (if we do)
+  - reset the running total and keep going
+- the last `ListNode` in the input list will have a `.next` of `None`, so we can iterate over the input list using a `while` loop with that as our stopping condition.
+  - but since that will break the loop before we can create the last `ListNode` from the current running total, we'll need to create the last one after the loop breaks
+
+## Refining the problem
+
+## Attempted solution(s)
+```python
+# 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]:
+        curr_input_node = head.next
+        curr_sum = 0
+        while curr_input_node.next is not None:
+            if curr_input_node.val == 0:
+                new_merged_node = ListNode(val=curr_sum)
+                try:
+                    prev_merged_node.next = new_merged_node
+                except NameError:
+                    new_head_node = new_merged_node
+                curr_sum = 0
+                prev_merged_node = new_merged_node
+            else:
+                curr_sum += curr_input_node.val
+
+            curr_input_node = curr_input_node.next
+
+        try:
+            prev_merged_node.next = ListNode(val=curr_sum)
+        except NameError:
+            new_head_node = ListNode(val=curr_sum)
+
+        return new_head_node
+```
+
+![](https://github.com/paxtonfitzpatrick/leetcode-solutions/assets/26118297/bd8a392e-113c-48de-a3f7-ae5217980630)
+
+That's kinda slow... there's definitely a more efficient way to do this; might revisit this one at some point.