Sentinel and two head pointers
Join Two Ordered Relays
01. The question
Build two nondecreasing singly linked chains from {"left":[...],"right":[...]}. Relink their existing nodes into one nondecreasing chain and return its readings. On equal values take the left node first.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Input{"left":[1,4,4],"right":[2,4,7]}
Output[1,2,4,4,4,7]
- Attach left’s 1 after a sentinel, then right’s 2.
- The two left 4 nodes win ties against right’s 4.
- When left is exhausted, attach the remaining right chain.
Try your code
Define solve(data, parameter="") and return your result. Your code runs locally, not on a server.
Edit your test cases
Use 1-10 objects with data and an optional string parameter. Expected outputs are computed from the lesson. An optional expected field must be correct.
Input contract & reference code
Example data: {"left":[1,4,4],"right":[2,4,7]}. Parameter: "". Numeric parameters arrive as strings; convert them with int(parameter) when needed.
The example and generated cases use the question’s exact variant. This checks behavior on a small test batch—not performance on every possible input.
Python downloads only when you run code, using compressed runtime assets. Code is not saved or uploaded. Enable JavaScript to run it.
Printed output
Five-second execution limit per batch. Standard-library Python only; additional packages and network access are unavailable. Stop ends execution without blocking the lesson.
03. The idea
Try sketching your own approach first. What would you track, and why would it work?
Reveal the idea
- Use a dummy node so the first append needs no special case.
- Attach the smaller head and advance only its source pointer.
- Move the output tail; finally link the entire remaining source.
04. Watch it happen
Use Step to move slowly, or Play to follow the whole trace. Change a preset or load your own input.
Input & settings
Red = current item / blue = active window or cells / green = completed work. A value docks at its actual destination; the original input stays unchanged. Positions start at 0.
Working state / what changes
Ready to begin
Enable JavaScript to step through the animation. The worked example and solution below remain readable without it.
Python / why it changes
The line behind the step
class Node:
def __init__(self, value):
self.value = value
self.next = None
def chain(values):
dummy = Node(None)
tail = dummy
for value in values:
tail.next = Node(value)
tail = tail.next
return dummy.next
def values(head):
result = []
while head is not None:
result.append(head.value)
head = head.next
return result
def solve(data, parameter=''):
left, right = chain(data["left"]), chain(data["right"])
dummy = Node(None)
tail = dummy
while left is not None and right is not None:
if left.value <= right.value:
tail.next = left
left = left.next
else:
tail.next = right
right = right.next
tail = tail.next
tail.next = left if left is not None else right
return values(dummy.next)How to use this walkthrough
Hover or focus a control for its label. Play forms the input and follows values into the working state. Step shows one change; Back and the slider restore exact snapshots. Reduced motion uses instant updates. Saved snapshots and sorted heap displays are teaching overhead, not algorithm complexity.
05. Python solution
class Node:
def __init__(self, value):
self.value = value
self.next = None
def chain(values):
dummy = Node(None)
tail = dummy
for value in values:
tail.next = Node(value)
tail = tail.next
return dummy.next
def values(head):
result = []
while head is not None:
result.append(head.value)
head = head.next
return result
def solve(data, parameter=''):
left, right = chain(data["left"]), chain(data["right"])
dummy = Node(None)
tail = dummy
while left is not None and right is not None:
if left.value <= right.value:
tail.next = left
left = left.next
else:
tail.next = right
right = right.next
tail = tail.next
tail.next = left if left is not None else right
return values(dummy.next)06. Cost & edge cases
- Time
- O(n + m)
- Space
- O(1) algorithm workspace; O(n + m) boundary construction and returned JSON
- Either or both input chains may be empty.
- Repeated readings are retained.
- Unsorted inputs are rejected rather than silently sorted.