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

↓

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 C++

1. Create min heap
2. Insert first nodes of lists
3. While heap not empty:
4.   Extract min node
5.   Add to merged list
6.   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 After Insertion
1	Insert first nodes	[1, 2, 3]	None	Empty	[1, 2, 3]
2	Extract min	[1, 2, 3]	1	1 -> null	[2, 3]
3	Insert next of extracted (1->4)	[2, 3]	None	1 -> null	[2, 3, 4]
4	Extract min	[2, 3, 4]	2	1 -> 2 -> null	[3, 4]
5	Insert next of extracted (2->6)	[3, 4]	None	1 -> 2 -> null	[3, 4, 6]
6	Extract min	[3, 4, 6]	3	1 -> 2 -> 3 -> null	[4, 6]
7	Insert next of extracted (3->5)	[4, 6]	None	1 -> 2 -> 3 -> null	[4, 5, 6]
8	Extract min	[4, 5, 6]	4	1 -> 2 -> 3 -> 4 -> null	[5, 6]
9	Insert next of extracted (4->null)	[5, 6]	None	1 -> 2 -> 3 -> 4 -> null	[5, 6]
10	Extract min	[5, 6]	5	1 -> 2 -> 3 -> 4 -> 5 -> null	[6]
11	Insert next of extracted (5->null)	[6]	None	1 -> 2 -> 3 -> 4 -> 5 -> null	[6]
12	Extract min	[6]	6	1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null	[]
13	Heap empty, done	[]	None	1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null	[]

💡 Heap is empty, all nodes merged

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	After Step 8	After Step 10	Final
Heap	[]	[2, 3]	[3, 4]	[4, 6]	[5, 6]	[6]	[]
Merged List	null	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

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 smallest candidate. This keeps the heap size manageable and ensures correct order, as shown in steps 3, 5, 7, 9, and 11 in the execution_table.

What happens if one of the lists is empty at the start?

If a list is empty, we simply don't insert any node from it into the heap initially. The heap only contains nodes from non-empty lists, so the algorithm still works correctly without that list, as implied in step 1.

How does the heap help in merging k sorted lists efficiently?

The heap always gives the smallest current node among all lists in O(log k) time. This avoids scanning all lists repeatedly. The execution_table shows how the heap content changes and always extracts the smallest node.

Visual Quiz - 3 Questions

Look at the execution_table at Step 4. What is the merged list state after extracting the min node?

A 1 -> 2 -> null

B 2 -> null

C 1 -> null

D Empty

💡 Hint

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

At which step does the heap become empty, ending the merge process?

A Step 12

B Step 13

C Step 10

D Step 8

💡 Hint

Look at the 'Heap Content' column and the exit_note in the execution_table.

If the next node of extracted node 3 was null, what would happen to the heap after Step 6?

A Heap would become [4, 6, 5]

B Heap would become [6]

C Heap would remain [4, 6]

D Heap would be empty

💡 Hint

Refer to Step 7 where the next node of 3 is inserted; if null, no insertion happens.

Concept Snapshot

Merge K Sorted Lists Using Min Heap:
- Insert first node of each list into a 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 is a fully merged sorted 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, we repeatedly extract the smallest node from the heap, add it to our merged list, and if that node has a next node, we insert it into the heap. This process continues until the heap is empty, meaning all nodes are merged. The heap helps efficiently find the smallest node among all lists at each step, keeping the merged list sorted. This method handles empty lists by simply not inserting nodes from them initially. The key is that each insertion into the heap is the next node from the list of the extracted node, ensuring order and efficiency.

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 list individually before merging.

C. It uses extra memory 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 quickly find the smallest element among multiple sorted lists by always keeping the smallest element at the top.

2. Step 2: Compare with other options

   Options A, B, and C do not describe the min heap's advantage correctly. The min heap does not sort lists individually or concatenate without sorting.

3. Final Answer:

   It efficiently finds the smallest element among the heads of all lists. -> 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
• Assuming min heap stores all elements at once
• Confusing min heap with simple concatenation

2. Which of the following is the correct way to push an element into a min heap in C++ using STL?

easy

A. std::priority_queue minHeap; minHeap.push(5);

B. std::priority_queue<int> minHeap; minHeap.push(5);

C. std::priority_queue minHeap; minHeap.push(5);

D. std::priority_queue minHeap; minHeap.push(5);

Solution

1. Step 1: Identify min heap declaration syntax

   In C++, to create a min heap, use std::priority_queue with std::greater comparator and std::vector as container.

2. Step 2: Check each option

   std::priority_queue minHeap; minHeap.push(5); uses std::greater and vector correctly. std::priority_queue<int> minHeap; minHeap.push(5); is max heap by default. std::priority_queue minHeap; minHeap.push(5); uses std::less which is max heap. std::priority_queue minHeap; minHeap.push(5); uses std::list which is invalid container for priority_queue.

3. Final Answer:

   std::priority_queue minHeap; minHeap.push(5); -> Option C

4. Quick Check:

   Min heap = priority_queue + greater + vector [OK]

Hint: Min heap needs std::greater and vector container [OK]

Common Mistakes:

• Using default priority_queue (max heap)
• Using std::less instead of std::greater
• Using invalid container like std::list

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

#include <iostream>
#include <vector>
#include <queue>

struct Node {
    int val;
    int listIndex;
    int elementIndex;
    Node(int v, int li, int ei) : val(v), listIndex(li), elementIndex(ei) {}
};

struct Compare {
    bool operator()(Node const& a, Node const& b) {
        return a.val > b.val;
    }
};

int main() {
    std::vector lists = {{1,4,5}, {1,3,4}, {2,6}};
    std::priority_queue minHeap;

    for (int i = 0; i < lists.size(); i++) {
        if (!lists[i].empty())
            minHeap.push(Node(lists[i][0], i, 0));
    }

    std::vector<int> merged;
    while (!minHeap.empty()) {
        Node curr = minHeap.top();
        minHeap.pop();
        merged.push_back(curr.val);
        int nextIndex = curr.elementIndex + 1;
        if (nextIndex < lists[curr.listIndex].size()) {
            minHeap.push(Node(lists[curr.listIndex][nextIndex], curr.listIndex, nextIndex));
        }
    }

    for (int val : merged) std::cout << val << " ";
    return 0;
}

medium

A. 1 1 2 3 4 4 5 6

B. 1 2 3 4 4 5 6

C. 1 1 3 4 4 5 6 2

D. 2 1 1 3 4 4 5 6

Solution

1. Step 1: Understand the merging process

   The code pushes the first element of each list into the min heap. Then it repeatedly extracts the smallest element and pushes the next element from the same list until all elements are merged.

2. Step 2: Trace the merged output

   Lists: {1,4,5}, {1,3,4}, {2,6}. Merged order: 1 (from list0), 1 (list1), 2 (list2), 3 (list1), 4 (list0), 4 (list1), 5 (list0), 6 (list2).

3. Final Answer:

   1 1 2 3 4 4 5 6 -> Option A

4. Quick Check:

   Merge sorted lists = 1 1 2 3 4 4 5 6 [OK]

Hint: Push first elements, pop min, push next from same list [OK]

Common Mistakes:

• Mixing order of elements
• Forgetting to push next element after pop
• Ignoring duplicates in merged output

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

std::priority_queue minHeap;
for (auto& list : lists) {
    for (int val : list) {
        minHeap.push(val);
    }
}
std::vector<int> merged;
while (!minHeap.empty()) {
    merged.push_back(minHeap.top());
    minHeap.pop();
}

medium

A. The min heap is declared incorrectly and will cause a compilation error.

B. The code pushes all elements at once, losing the benefit of merging sorted lists efficiently.

C. The code does not pop elements from the min heap, causing an infinite loop.

D. The merged vector is not initialized before use.

Solution

1. Step 1: Analyze the approach used

   The code pushes all elements from all lists into the min heap at once, then pops them to get sorted order.

2. Step 2: Identify inefficiency

   This approach loses the efficiency of merging K sorted lists using min heap by only pushing the first element of each list and pushing next elements on demand.

3. Final Answer:

   The code pushes all elements at once, losing the benefit of merging sorted lists efficiently. -> Option B

4. Quick Check:

   Push all elements at once = inefficient merge [OK]

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

Common Mistakes:

• Pushing all elements at once
• Ignoring heap declaration correctness
• Forgetting to pop elements in loop

5. Given K sorted linked lists, how can you modify the min heap approach to merge them efficiently if some lists are empty and others have duplicate values?

hard

A. Ignore empty lists and merge only non-empty lists by concatenation without using a min heap.

B. Use a max heap instead of min heap to handle duplicates and empty lists.

C. Push all elements from all lists into the min heap at once, then pop to merge, skipping duplicates manually.

D. Initialize the min heap with the first non-empty element of each list and push next elements only when popping, allowing duplicates naturally.

Solution

1. Step 1: Handle empty lists and duplicates

   Skip empty lists when initializing the min heap. For duplicates, min heap naturally handles them by ordering values.

2. Step 2: Efficient merging strategy

   Push only the first element of each non-empty list into the min heap. When popping, push the next element from the same list if it exists. This keeps memory and operations efficient.

3. Final Answer:

   Initialize the min heap with the first non-empty element of each list and push next elements only when popping, allowing duplicates naturally. -> Option D

4. Quick Check:

   Push first non-empty elements, push next on pop, duplicates allowed [OK]

Hint: Skip empty lists, push next element after pop [OK]

Common Mistakes:

• Pushing all elements at once wasting memory
• Ignoring empty lists causing errors
• Using max heap which breaks order