Merge K Sorted Lists Using Min Heap in DSA C - 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 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 each list, 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 C

1. Create min heap
2. Insert first nodes of lists
3. While heap not empty:
   - Extract min
   - Add to merged list
   - Insert next node if exists

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

Execution Table

Step	Operation	Heap Content (values)	Extracted Node	Merged List State	Heap Changes
1	Insert first nodes	[1, 2, 3]	None	Empty	Heap built with first nodes of each list
2	Extract min	[2, 3]	1	1 -> null	Removed 1 from heap
3	Insert next of extracted (1->4)	[2, 3, 4]	None	1 -> null	Inserted 4 into heap
4	Extract min	[3, 4]	2	1 -> 2 -> null	Removed 2 from heap
5	Insert next of extracted (2->6)	[3, 4, 6]	None	1 -> 2 -> null	Inserted 6 into heap
6	Extract min	[4, 6]	3	1 -> 2 -> 3 -> null	Removed 3 from heap
7	Insert next of extracted (3->5)	[4, 6, 5]	None	1 -> 2 -> 3 -> null	Inserted 5 into heap
8	Extract min	[5, 6]	4	1 -> 2 -> 3 -> 4 -> null	Removed 4 from heap
9	Insert next of extracted (4->7)	[5, 6, 7]	None	1 -> 2 -> 3 -> 4 -> null	Inserted 7 into heap
10	Extract min	[6, 7]	5	1 -> 2 -> 3 -> 4 -> 5 -> null	Removed 5 from heap
11	No next node for 5	[6, 7]	None	1 -> 2 -> 3 -> 4 -> 5 -> null	No insertion
12	Extract min	[7]	6	1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null	Removed 6 from heap
13	No next node for 6	[7]	None	1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null	No insertion
14	Extract min	[]	7	1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null	Removed 7 from heap
15	No next node for 7	[]	None	1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null	No insertion

💡 Heap is empty, all nodes merged

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 5	After Step 7	After Step 9	After Step 11	After Step 13	Final
Heap	Empty	[1,2,3]	[2,3,4]	[3,4,6]	[4,6,5]	[5,6,7]	[6,7]	[7]	[]
Merged List	Empty	Empty	1 -> null	1 -> 2 -> null	1 -> 2 -> 3 -> null	1 -> 2 -> 3 -> 4 -> null	1 -> 2 -> 3 -> 4 -> 5 -> null	1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null	1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null

Key Moments - 3 Insights

Why do we insert 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 might be the next smallest overall. This is shown in execution_table rows 3, 5, 7, 9 where next nodes are inserted to keep heap updated.

What happens when a list runs out of nodes?

When a list has no next node, we do not insert anything into the heap. This is shown in execution_table rows 11, 13, and 15 where no insertion happens after extracting nodes 5, 6, and 7.

Why does the heap start with the first nodes of all lists?

Because the first nodes are the smallest elements of each sorted list, so starting with them allows the heap to always pick the smallest among all lists at each step. This is shown in execution_table row 1.

Visual Quiz - 3 Questions

Look at the execution_table at Step 4, what is the merged list state?

A 2 -> null

B 1 -> null

C 1 -> 2 -> null

D Empty

💡 Hint

Check the 'Merged List State' column at Step 4 in execution_table

At which step does the heap become empty?

A Step 15

B Step 14

C Step 13

D Step 12

💡 Hint

Look at the 'Heap Content' column in execution_table rows 14 and 15

If the first node of the second list was 0 instead of 2, what would be the heap content after Step 1?

A [0, 1, 3]

B [1, 2, 3]

C [0, 2, 3]

D [1, 0, 3]

💡 Hint

Heap contains first nodes of each list; replacing 2 with 0 changes heap accordingly as shown in Step 1

Concept Snapshot

Merge K Sorted Lists Using Min Heap:
- Insert first node of each list into min heap
- Extract min node from heap and add to merged list
- Insert next node of extracted node into heap if exists
- Repeat until heap is empty
- Result: merged sorted linked list

Full Transcript

To merge k sorted linked 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, while the heap is not empty, we extract the smallest node, add it to the merged list, and if the extracted node has a next node, we insert that next node into the heap. This process continues until all nodes are merged. The heap helps efficiently find the smallest node among all lists at each step.

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 all lists at each step.

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 without sorting.

Solution

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

   A min heap helps to quickly get the smallest element among multiple sorted lists without scanning all elements.

2. Step 2: Compare options based on heap usage

   Only It efficiently finds the smallest element among all lists at each step. correctly states the advantage of min heap in efficiently finding the smallest element at each step.

3. Final Answer:

   It efficiently finds the smallest element among all lists at each step. -> Option A

4. Quick Check:

   Min heap = efficient smallest element retrieval [OK]

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

Common Mistakes:

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

2. Which of the following is the correct way to initialize a min heap in C for merging K sorted lists?

easy

A. heap<int> minHeap = make_heap();

B. minHeap = new MinHeap();

C. priority_queue minHeap;

D. int minHeap[] = {};

Solution

1. Step 1: Recognize C++ STL syntax for min heap

   In C++, a min heap is created using priority_queue with greater comparator as in priority_queue minHeap;.

2. Step 2: Check other options for correctness

   minHeap = new MinHeap(); is incomplete syntax, C is invalid, and D is just an array, not a heap.

3. Final Answer:

   priority_queue minHeap; -> Option C

4. Quick Check:

   Min heap init in C++ STL = priority_queue with greater comparator [OK]

Hint: Use priority_queue with greater<int> for min heap [OK]

Common Mistakes:

• Using default priority_queue (max heap)
• Trying to create heap with invalid syntax
• Confusing array with heap structure

3. Given three sorted lists: [1,4,5], [1,3,4], and [2,6], what is the merged sorted list using a min heap?

medium

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

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

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

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

Solution

1. Step 1: Combine all elements maintaining order

   All elements from lists combined: 1,4,5; 1,3,4; 2,6.

2. Step 2: Sort combined elements

   Sorted combined list is [1,1,2,3,4,4,5,6].

3. Final Answer:

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

4. Quick Check:

   Merge sorted lists = sorted combined elements [OK]

Hint: Merge all, then sort or use min heap to pick smallest [OK]

Common Mistakes:

• Missing duplicate 1 or 4
• Skipping 2 or 3 in merged list
• Incorrect order of elements

4. What is the error in the following code snippet for merging K sorted lists using a min heap?

for (int i = 0; i < lists.size(); i++) {
    minHeap.push(lists[i][0]);
}
while (!minHeap.empty()) {
    int val = minHeap.top();
    minHeap.pop();
    result.push_back(val);
}

medium

A. It uses pop() before top(), causing runtime error.

B. It only pushes the first element of each list and never pushes next elements.

C. It does not initialize the min heap properly.

D. It pushes all elements at once, causing overflow.

Solution

1. Step 1: Analyze the push operation

   The code pushes only the first element of each list into the min heap.

2. Step 2: Check if next elements are pushed

   There is no code to push the next elements from lists after popping the smallest element.

3. Final Answer:

   It only pushes the first element of each list and never pushes next elements. -> Option B

4. Quick Check:

   Missing pushing next elements after pop = bug [OK]

Hint: Always push next element from list after popping min [OK]

Common Mistakes:

• Not pushing next elements after popping
• Popping before checking top
• Incorrect heap initialization

5. You have K sorted linked lists of varying lengths. Which approach best optimizes merging them using a min heap to achieve O(N log K) time complexity, where N is total elements?

hard

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

B. Concatenate all lists into one and then sort the combined list using quicksort.

C. Push all nodes from all lists into the min heap at once, then pop all to get sorted order.

D. Merge lists two at a time using simple merge sort repeatedly until one list remains.

Solution

1. Step 1: Understand time complexity of each approach

   Push the first node of each list into the min heap, then repeatedly pop the smallest and push its next node until all nodes are processed. uses a min heap of size K, pushing and popping nodes efficiently, achieving O(N log K).

2. Step 2: Compare other options

   Concatenate all lists into one and then sort the combined list using quicksort. sorts all elements at once, O(N log N). Push all nodes from all lists into the min heap at once, then pop all to get sorted order. pushes all nodes at once, O(N log N). Merge lists two at a time using simple merge sort repeatedly until one list remains. merges lists pairwise, O(N log K) but less efficient than min heap approach.

3. Final Answer:

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

4. Quick Check:

   Min heap with K nodes = O(N log K) merge [OK]

Hint: Push first nodes, pop min, push next nodes for O(N log K) [OK]

Common Mistakes:

• Pushing all nodes at once causing O(N log N)
• Using pairwise merges inefficiently
• Sorting concatenated list instead of merging