Merge K Sorted Lists Using Min Heap in DSA Go - 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 min node to merged list

↓

If min node has next, insert next into heap

↓

Repeat until heap 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 Go

lists = [[1,4,5],[1,3,4],[2,6]]
minHeap = []
for each list:
  insert first node into minHeap
while minHeap not empty:
  node = extract min
  add node to merged list
  if node.next:
    insert node.next into minHeap

This code merges 3 sorted lists by always picking the smallest current node using a min heap.

Execution Table

Step	Operation	Nodes in Heap	Extracted Node	Heap After Extract	Merged List State
1	Insert first nodes	[1(L1),1(L2),2(L3)]	None	[1(L1),1(L2),2(L3)]	Empty
2	Extract min	[1(L1),1(L2),2(L3)]	1(L1)	[1(L2),2(L3)]	1
3	Insert next of extracted	[1(L2),2(L3),4(L1)]	None	[1(L2),2(L3),4(L1)]	1
4	Extract min	[1(L2),2(L3),4(L1)]	1(L2)	[2(L3),4(L1)]	1 -> 1
5	Insert next of extracted	[2(L3),3(L2),4(L1)]	None	[2(L3),3(L2),4(L1)]	1 -> 1
6	Extract min	[2(L3),3(L2),4(L1)]	2(L3)	[3(L2),4(L1)]	1 -> 1 -> 2
7	Insert next of extracted	[3(L2),4(L1),6(L3)]	None	[3(L2),4(L1),6(L3)]	1 -> 1 -> 2
8	Extract min	[3(L2),4(L1),6(L3)]	3(L2)	[4(L1),6(L3)]	1 -> 1 -> 2 -> 3
9	Insert next of extracted	[4(L1),4(L2),6(L3)]	None	[4(L1),4(L2),6(L3)]	1 -> 1 -> 2 -> 3
10	Extract min	[4(L1),4(L2),6(L3)]	4(L1)	[4(L2),6(L3)]	1 -> 1 -> 2 -> 3 -> 4
11	Insert next of extracted	[4(L2),5(L1),6(L3)]	None	[4(L2),5(L1),6(L3)]	1 -> 1 -> 2 -> 3 -> 4
12	Extract min	[4(L2),5(L1),6(L3)]	4(L2)	[5(L1),6(L3)]	1 -> 1 -> 2 -> 3 -> 4 -> 4
13	Insert next of extracted	[5(L1),6(L3)]	None	[5(L1),6(L3)]	1 -> 1 -> 2 -> 3 -> 4 -> 4
14	Extract min	[5(L1),6(L3)]	5(L1)	[6(L3)]	1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5
15	Insert next of extracted	[6(L3)]	None	[6(L3)]	1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5
16	Extract min	[6(L3)]	6(L3)	[]	1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6
17	Insert next of extracted	[]	None	[]	1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6
18	Heap empty	[]	None	[]	Merged list complete

💡 Heap is empty, all nodes merged into one sorted list

Variable Tracker

Variable	Start	After Step 1	After Step 5	After Step 10	After Step 15	Final
minHeap	[]	[1(L1),1(L2),2(L3)]	[2(L3),3(L2),4(L1)]	[4(L2),6(L3)]	[6(L3)]	[]
mergedList	Empty	Empty	1 -> 1	1 -> 1 -> 2 -> 3 -> 4	1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5	1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6

Key Moments - 3 Insights

Why do we insert only the first node of each list into the heap initially?

Because the lists are sorted, the first node is the smallest in each list. We start with these to find the overall smallest node among all lists, as shown in execution_table step 1.

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

After extracting the smallest node, the next node in that list might be the next smallest candidate. Inserting it keeps the heap updated with current smallest nodes, as seen in steps 3, 5, 7, etc.

How do we know when to stop the merging process?

When the heap becomes empty, it means all nodes from all lists have been extracted and merged. This is shown in execution_table step 18.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, which node was extracted from the heap?

A 1 from list 1

B 1 from list 2

C 2 from list 3

D 4 from list 1

💡 Hint

Check the 'Extracted Node' column at step 4 in execution_table

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

A Step 6

B Step 10

C Step 8

D Step 12

💡 Hint

Look at the 'Merged List State' column and find when '3' appears

If we did not insert the next node after extraction, what would happen to the heap size over time?

A Heap size would shrink and eventually become empty too soon

B Heap size would stay constant

C Heap size would grow indefinitely

D Heap size would alternate between 1 and 2

💡 Hint

Refer to variable_tracker for minHeap size changes after insertions

Concept Snapshot

Merge K Sorted Lists Using Min Heap:
- Insert first node of each list into min heap
- Extract min node from heap, add to merged list
- Insert next node of extracted node into heap if exists
- 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. We start by inserting the first node of each list into the heap. Then, we repeatedly extract the smallest node, add it to the merged list, and insert the next node from that node's list into the heap if it exists. This process continues until the heap is empty, meaning all nodes are merged. The execution table shows each step's heap state, extracted node, and merged list state. The variable tracker follows the heap and merged list changes. Key moments clarify why we insert only the first nodes initially, why we insert next nodes after extraction, and when the process stops. The visual quiz tests understanding of these steps. 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 uses extra memory to store all elements at once.

B. It sorts each list individually before merging.

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

D. It merges lists by concatenating them without sorting.

Solution

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

   A min heap keeps track of the smallest elements efficiently, allowing quick access to the smallest head among K lists.

2. Step 2: Compare other options

   Options B, C, and D do not describe the min heap's role correctly. Sorting individually or concatenating without sorting does not merge sorted lists efficiently.

3. Final Answer:

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

4. Quick Check:

   Min heap = efficient smallest element access [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 min heap with simple concatenation

2. Which Go data structure is best suited to implement a min heap for merging K sorted lists?

easy

A. map

B. linked list

C. array without heap functions

D. slice with container/heap package

Solution

1. Step 1: Identify Go's min heap implementation

   Go uses slices combined with the container/heap package to implement min heaps efficiently.

2. Step 2: Evaluate other options

   Maps and linked lists do not provide heap properties. Arrays without heap functions lack automatic heap behavior.

3. Final Answer:

   slice with container/heap package -> Option D

4. Quick Check:

   Go min heap = slice + container/heap [OK]

Hint: Use container/heap with slice for min heap in Go [OK]

Common Mistakes:

• Using map which has no order
• Using linked list which is slow for heap ops
• Ignoring container/heap package

3. What is the output of merging these 3 sorted lists using a min heap?

list1 = [1, 4, 7]
list2 = [2, 5, 8]
list3 = [3, 6, 9]

medium

A. [1, 3, 5, 7, 9, 2, 4, 6, 8]

B. [1, 2, 3, 4, 5, 6, 7, 8, 9]

C. [9, 8, 7, 6, 5, 4, 3, 2, 1]

D. [1, 4, 7, 2, 5, 8, 3, 6, 9]

Solution

1. Step 1: Understand merging with min heap

   The min heap picks the smallest head element from all lists repeatedly, producing a fully sorted merged list.

2. Step 2: Merge given lists step-by-step

   Starting with heads: 1, 2, 3; pick 1, then 2, then 3, then 4, 5, 6, 7, 8, 9 in order.

3. Final Answer:

   [1, 2, 3, 4, 5, 6, 7, 8, 9] -> Option B

4. Quick Check:

   Merge sorted lists = sorted combined list [OK]

Hint: Min heap merges by always picking smallest next element [OK]

Common Mistakes:

• Concatenating lists without sorting
• Reversing order incorrectly
• Mixing elements from different lists randomly

4. Given this Go snippet for merging K sorted lists, what is the bug?

func mergeKLists(lists [][]int) []int {
    var result []int
    h := &IntHeap{}
    heap.Init(h)
    for _, list := range lists {
        for _, val := range list {
            heap.Push(h, val)
        }
    }
    for h.Len() > 0 {
        result = append(result, heap.Pop(h).(int))
    }
    return result
}

medium

A. Pushing all elements at once loses the benefit of merging sorted lists.

B. Heap is not initialized before pushing elements.

C. Result slice is not declared properly.

D. Heap.Pop returns wrong type.

Solution

1. Step 1: Analyze the merging approach

   The code pushes all elements from all lists into the heap at once, ignoring the sorted order advantage.

2. Step 2: Understand the impact

   This approach works but is inefficient (O(N log N)) instead of O(N log K) by pushing only heads and updating heap dynamically.

3. Final Answer:

   Pushing all elements at once loses the benefit of merging sorted lists. -> Option A

4. Quick Check:

   Push all elements = inefficient merge [OK]

Hint: Push only heads, not all elements at once [OK]

Common Mistakes:

• Ignoring heap.Init call
• Misunderstanding result slice declaration
• Incorrect type assertion on Pop

5. You have K sorted linked lists of varying lengths. How would you efficiently merge them using a min heap in Go to minimize time complexity?

hard

A. Push the first node of each list into the min heap, then repeatedly pop the smallest and push its next node.

B. Push all nodes from all lists into the min heap at once, then pop all to get merged list.

C. Concatenate all lists and sort the combined list after.

D. Use a max heap instead of a min heap for merging.

Solution

1. Step 1: Use min heap with heads of each list

   Push only the first node of each list into the min heap to keep track of the smallest current element among lists.

2. Step 2: Pop smallest and push next node

   Pop the smallest node, add it to the merged list, then push the next node from that list into the heap, repeating until all nodes are processed.

3. Step 3: Time complexity advantage

   This approach runs in O(N log K) time, where N is total nodes and K is number of lists, which is efficient.

4. Final Answer:

   Push the first node of each list into the min heap, then repeatedly pop the smallest and push its next node. -> Option A

5. Quick Check:

   Push heads + pop/push next = efficient merge [OK]

Hint: Push heads only, then push next nodes dynamically [OK]

Common Mistakes:

• Pushing all nodes at once causing inefficiency
• Using max heap which gives largest element first
• Sorting concatenated list which is slower