Heap of source heads
Join Several Ordered Relays
01. The question
Build a chain for every row in a JSON list of sorted integer lists. Relink nodes into one sorted chain and return its readings. Heap ties use source index then original node position for deterministic identity order.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Input[[1,6],[1,3,8],[],[2,6]]
Output[1,1,2,3,6,6,8]
- The heap holds one candidate head for each nonempty source.
- Pop source 0’s 1 before source 1’s equal 1.
- After attaching a node, offer its saved successor to the heap.
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: [[1,6],[1,3,8],[],[2,6]]. 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
- Push each nonempty source head into a min-heap.
- Pop the next smallest node, save its successor and detach it into the output.
- Push that successor; stop when the heap is empty.
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
from heapq import heappush, heappop
def solve(data, parameter=''):
heap = []
for source, readings in enumerate(data):
head = chain(readings)
if head is not None:
heappush(heap, (head.value, source, 0, head))
dummy = Node(None)
tail = dummy
while heap:
value, source, position, node = heappop(heap)
following = node.next
tail.next = node
tail = node
tail.next = None
if following is not None:
heappush(heap, (following.value, source, position + 1, following))
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
from heapq import heappush, heappop
def solve(data, parameter=''):
heap = []
for source, readings in enumerate(data):
head = chain(readings)
if head is not None:
heappush(heap, (head.value, source, 0, head))
dummy = Node(None)
tail = dummy
while heap:
value, source, position, node = heappop(heap)
following = node.next
tail.next = node
tail = node
tail.next = None
if following is not None:
heappush(heap, (following.value, source, position + 1, following))
return values(dummy.next)06. Cost & edge cases
- Time
- O(k + N log(max(2, k)))
- Space
- O(k) algorithm workspace; O(N) boundary nodes and returned JSON
- An empty source list or only empty rows returns [].
- Heap pictures show next-pop order, not the heap’s internal array.
- Every row must already be nondecreasing.