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?

A min heap helps efficiently find the smallest element among the heads of all lists. We repeatedly extract the smallest element and add the next element from the same list to the heap until all lists are merged.

intermediate

What is the time complexity of merging K sorted lists using a min heap?

The time complexity is O(N log K), where N is the total number of elements across all lists, and K is the number of lists. This is because each element is pushed and popped from the heap which has size at most K.

beginner

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

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

intermediate

Why is a min heap preferred over a simple linear search when merging K sorted lists?

A min heap reduces the search for the smallest element from O(K) to O(log K) time, making the merging process much faster especially when K is large.

beginner

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

After extracting the smallest element, the next element from the same list is inserted into the min heap if it exists. This keeps the heap updated with the current smallest candidates.

What data structure is used to efficiently merge K sorted lists?

A Min heap

B Stack

C Queue

D Hash map

A min heap efficiently keeps track of the smallest elements among the K lists.

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

A N (total elements)

B N/K

C 1

D K (number of lists)

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

What is the time complexity of inserting or removing an element from the min heap?

A O(K)

B O(log K)

C O(1)

D O(N)

Insertion and removal in a min heap take O(log K) time where K is the heap size.

After extracting the smallest element from the min heap, what is the next step?

A Insert the next element from the same list into the heap

B Insert the largest element from any list

C Remove all elements from the heap

D Stop the process

We insert the next element from the same list to keep the heap updated.

If there are K lists each with N elements, what is the total number of elements merged?

A K + N

B N

C K * N

D K

Total elements merged is K multiplied by N.

Explain how a min heap helps merge K sorted lists efficiently.

Think about how to always get the smallest next element quickly.

Describe the steps involved in merging K sorted lists using a min heap.

Imagine merging by always picking the smallest available element.

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