Merge K Sorted Lists Using Min Heap in DSA Python - Execution Trace

Concept Flow - Merge K Sorted Lists Using Min Heap

Create min heap

↓

Insert first node of each list into heap

↓

While heap not empty

↓

Extract min node from heap

↓

Add extracted node to merged list

↓

If extracted node has next, insert next into heap

↓

While heap not empty

↓

Return merged list head

We build a min heap with the first nodes of all lists, then repeatedly extract the smallest node, add it to the merged list, and insert the next node from that list into the heap until all nodes are merged.

Execution Sample

DSA Python

lists = [[1,4,5],[1,3,4],[2,6]]
# Merge these using min heap
# Result: 1->1->2->3->4->4->5->6->null

This code merges three sorted lists into one sorted list using a min heap.

Execution Table

Step	Operation	Heap Content (values)	Extracted Node	Merged List State	Heap After Insertion
1	Insert first nodes	[1,1,2]	None	Empty	[1,1,2]
2	Extract min	[1,2]	1 (from list 1)	1	[1,2,4] (insert next 4 from list 1)
3	Extract min	[2,4]	1 (from list 2)	1->1	[2,3,4] (insert next 3 from list 2)
4	Extract min	[3,4]	2 (from list 3)	1->1->2	[3,4,6] (insert next 6 from list 3)
5	Extract min	[4,6]	3 (from list 2)	1->1->2->3	[4,4,6] (insert next 4 from list 2)
6	Extract min	[4,6]	4 (from list 1)	1->1->2->3->4	[4,5,6] (insert next 5 from list 1)
7	Extract min	[5,6]	4 (from list 2)	1->1->2->3->4->4	[5,6] (list 2 exhausted, no insertion)
8	Extract min	[6]	5 (from list 1)	1->1->2->3->4->4->5	[6] (list 1 exhausted, no insertion)
9	Extract min	[]	6 (from list 3)	1->1->2->3->4->4->5->6	[] (list 3 exhausted, no insertion)
10	Heap empty	[]	None	Merged list complete	[]

💡 Heap is empty, all nodes merged into the final list.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	After Step 9	Final
Heap	[]	[1,1,2]	[1,2,4]	[2,3,4]	[3,4,6]	[4,4,6]	[4,5,6]	[5,6]	[6]	[]	[]
Merged List	Empty	Empty	1	1->1	1->1->2	1->1->2->3	1->1->2->3->4	1->1->2->3->4->4	1->1->2->3->4->4->5	1->1->2->3->4->4->5->6	1->1->2->3->4->4->5->6

Key Moments - 3 Insights

Why do we insert only the next node of the extracted node into the heap?

Because each list is sorted, after extracting the smallest node, the next node in that list is the next candidate to merge. This keeps the heap size manageable and ensures correct order, as shown in steps 2, 3, 4, etc. in the execution_table.

What happens when a list is exhausted (no next node)?

No new node is inserted into the heap from that list, reducing heap size by one. This is seen in steps 7, 8, and 9 where no insertion happens after extracting the last nodes.

Why does the heap always contain at most k nodes?

Because we start with one node from each of the k lists and only insert the next node from the list of the extracted node. So heap size never exceeds k, as seen in the heap content column.

Visual Quiz - 3 Questions

Look at the execution_table at Step 4, what is the heap content after inserting the next node?

A [2,3,4]

B [3,4,6]

C [4,5,6]

D [1,2,4]

💡 Hint

Check the 'Heap After Insertion' column at Step 4 in the execution_table.

At which step does the merged list first contain the value 3?

A Step 5

B Step 3

C Step 4

D Step 6

💡 Hint

Look at the 'Merged List State' column in the execution_table to see when 3 appears.

If the first node of the third list was 0 instead of 2, how would the heap content change at Step 1?

A [0,1,2]

B [1,1,2]

C [0,1,1]

D [1,2,3]

💡 Hint

The heap starts with the first node of each list; changing the third list's first node to 0 affects the initial heap.

Concept Snapshot

Merge K Sorted Lists Using Min Heap:
- Insert first node of each list into a min heap
- Extract min node, add to merged list
- Insert next node from extracted node's list into heap
- Repeat until heap empty
- Result is one merged sorted list

Full Transcript

To merge k sorted lists, we use a min heap to always pick the smallest current node among the lists. We start by inserting the first node of each list into the heap. Then, we repeatedly extract the smallest node from the heap and add it to our merged list. If the extracted node has a next node, we insert that next node into the heap. This process continues until the heap is empty, meaning all nodes are merged. The heap size never exceeds k because we only keep one node from each list at a time. This method efficiently merges all lists into one sorted list.

Practice

1. What is the main advantage of using a min heap to merge k sorted lists?

easy

A. It efficiently finds the smallest element among the heads of all lists.

B. It sorts each individual list before merging.

C. It uses extra space to store all elements at once.

D. It merges lists by concatenating them directly.

Solution

1. Step 1: Understand the role of min heap in merging

   A min heap helps keep track of the smallest element among multiple sorted lists efficiently.

2. Step 2: Compare with other options

   Options B, C, and D do not describe the min heap's main advantage in this context.

3. Final Answer:

   It efficiently finds the smallest element among the heads of all lists. -> Option A

4. Quick Check:

   Min heap = smallest element tracking [OK]

Hint: Min heap always gives smallest element fast [OK]

Common Mistakes:

• Thinking min heap sorts entire lists
• Assuming min heap stores all elements at once
• Confusing concatenation with merging

2. Which of the following Python code snippets correctly initializes a min heap with the first elements of each sorted list named lists?

easy

A. heap = [lst[0] for lst in lists if lst]; import heapq; heapq.heapify(heap)

B. heap = lists[0]; import heapq; heapq.heapify(heap)

C. heap = []; for lst in lists: heap.append(lst); import heapq; heapq.heapify(heap)

D. heap = lists; import heapq; heapq.heapify(heap)

Solution

1. Step 1: Extract first elements safely

   heap = [lst[0] for lst in lists if lst]; import heapq; heapq.heapify(heap) uses list comprehension to get first elements only if the list is not empty.

2. Step 2: Check heap initialization

   heap = [lst[0] for lst in lists if lst]; import heapq; heapq.heapify(heap) then calls heapq.heapify to create a min heap from these elements correctly.

3. Final Answer:

   heap = [lst[0] for lst in lists if lst]; import heapq; heapq.heapify(heap) -> Option A

4. Quick Check:

   Correct min heap initialization = heap = [lst[0] for lst in lists if lst]; import heapq; heapq.heapify(heap) [OK]

Hint: Use list comprehension with condition and heapq.heapify [OK]

Common Mistakes:

• Not checking if list is empty before accessing lst[0]
• Heapifying entire lists instead of first elements
• Assigning heap to a single list instead of elements

3. What will be the output of the following code snippet that merges 3 sorted lists using a min heap?

import heapq
lists = [[1,4,5],[1,3,4],[2,6]]
heap = []
for i, lst in enumerate(lists):
    if lst:
        heapq.heappush(heap, (lst[0], i, 0))
result = []
while heap:
    val, list_idx, element_idx = heapq.heappop(heap)
    result.append(val)
    if element_idx + 1 < len(lists[list_idx]):
        next_tuple = (lists[list_idx][element_idx+1], list_idx, element_idx+1)
        heapq.heappush(heap, next_tuple)
print(result)

medium

A. [1, 2, 3, 4, 5, 6]

B. [1, 1, 2, 3, 4, 5, 6, 7]

C. [1, 1, 3, 4, 4, 5, 6]

D. [1, 1, 2, 3, 4, 4, 5, 6]

Solution

1. Step 1: Trace heap operations

   Initially, heap contains (1,0,0), (1,1,0), (2,2,0). We pop smallest 1, add next element 4 from list 0, then pop next 1, add 3 from list 1, and so on.

2. Step 2: Build merged list step-by-step

   Result builds as [1,1,2,3,4,4,5,6] by always popping smallest and pushing next from same list.

3. Final Answer:

   [1, 1, 2, 3, 4, 4, 5, 6] -> Option D

4. Quick Check:

   Merged sorted list = [1,1,2,3,4,4,5,6] [OK]

Hint: Track heap pops and pushes carefully [OK]

Common Mistakes:

• Missing duplicate 1s in output
• Forgetting to push next element from same list
• Adding elements out of order

4. Identify the error in the following code snippet that attempts to merge k sorted lists using a min heap:

import heapq
lists = [[1,3,5],[2,4,6],[0,7,8]]
heap = []
for lst in lists:
    heapq.heappush(heap, lst[0])
result = []
while heap:
    val = heapq.heappop(heap)
    result.append(val)
print(result)

medium

A. The lists variable is not defined.

B. The heap is not initialized using heapq.heapify().

C. The code pushes only first elements and never pushes next elements from lists.

D. The code uses heappop instead of pop.

Solution

1. Step 1: Analyze heap push operations

   The code pushes only the first element of each list into the heap once and never pushes subsequent elements.

2. Step 2: Understand why this causes incomplete merge

   Without pushing next elements, only first elements are merged, so the result is incomplete and incorrect.

3. Final Answer:

   The code pushes only first elements and never pushes next elements from lists. -> Option C

4. Quick Check:

   Missing next elements push = The code pushes only first elements and never pushes next elements from lists. [OK]

Hint: Always push next elements after popping from heap [OK]

Common Mistakes:

• Assuming heapq.heapify is mandatory here
• Confusing heappop with pop
• Ignoring the need to push next elements

5. Given k sorted linked lists, which approach best describes how to merge them efficiently using a min heap? A) Insert all nodes from all lists into a min heap, then pop all to form merged list. B) Push only the first node of each list into the min heap, then repeatedly pop the smallest node and push its next node until all nodes are merged. C) Merge lists two at a time using simple merge, then merge results. D) Concatenate all lists and sort the combined list.

hard

A. Insert all nodes at once into min heap and pop all.

B. Push first nodes, pop smallest, push next nodes repeatedly.

C. Merge lists two at a time without heap.

D. Concatenate all lists and sort at the end.

Solution

1. Step 1: Understand efficient min heap usage

   Pushing only the first node of each list keeps the heap small and efficient.

2. Step 2: Describe the merging process

   Pop smallest node, add it to merged list, then push the next node from that list to heap, repeating until done.

3. Step 3: Compare with other options

   Insert all nodes at once into min heap and pop all. wastes space and time by pushing all nodes at once. Options B and D do not use min heap.

4. Final Answer:

   Push first nodes, pop smallest, push next nodes repeatedly. -> Option B

5. Quick Check:

   Efficient merge = push first nodes, pop smallest, push next [OK]

Hint: Push first nodes only, then push next after popping [OK]

Common Mistakes:

• Pushing all nodes at once wastes memory
• Not pushing next nodes after popping
• Using simple merge instead of heap for k lists