Merge K Sorted Lists Using Min Heap in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind using a min heap to merge K sorted lists?

Use a min heap to always extract the smallest element among the heads of all lists, then insert the next element from that list into the heap, repeating until all lists are merged.

beginner

What data structure is typically used to efficiently find the smallest element among multiple sorted lists during merging?

A min heap (priority queue) is used because it allows quick access to the smallest element and efficient insertion of new elements.

intermediate

In the context of merging K sorted lists, what does each element in the min heap represent?

Each element in the min heap represents the current smallest node from one of the K lists, along with information about which list it came from.

intermediate

Why is the time complexity of merging K sorted lists using a min heap O(N log K)?

Because each of the N total elements is pushed and popped from the min heap, which has size at most K, and heap operations take O(log K) time.

beginner

What happens after extracting the smallest element from the min heap during the merge process?

The next element from the same list as the extracted element is inserted into the min heap if it exists, maintaining the heap property.

What is the initial step when merging K sorted lists using a min heap?

A Insert all elements of all lists into the min heap

B Concatenate all lists and sort them

C Insert the first element of each list into the min heap

D Merge two lists at a time

We start by inserting the first element of each list into the min heap to efficiently find the smallest among them.

What is the maximum size of the min heap during the merge of K sorted lists?

A K (number of lists)

B N/K

C 1

D N (total number of elements)

The min heap contains at most one element from each list at any time, so its size is at most K.

Which operation is repeated until all elements are merged in the min heap approach?

A Extract min and insert next element from the same list

B Extract max and insert previous element

C Sort the heap after each insertion

D Remove duplicates from the heap

We repeatedly extract the smallest element and insert the next element from the same list to maintain the heap.

What is the overall time complexity of merging K sorted lists with total N elements using a min heap?

A O(N log N)

B O(N log K)

C O(K log N)

D O(N + K)

Each of the N elements is pushed and popped from a heap of size K, so the complexity is O(N log K).

What data is stored in the min heap nodes during the merge?

A The entire list

B Only the element value

C Only the index of the list

D Element value and index of the list it belongs to

We store the element value and the list index to know where to get the next element after extraction.

Explain step-by-step how to merge K sorted lists using a min heap.

Think about how the min heap helps pick the smallest element at each step.

Describe why a min heap is efficient for merging K sorted lists compared to other methods.

Compare with sorting all elements or merging lists one by one.

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