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 current element among all lists by always keeping the smallest element at the top. This allows merging K sorted lists by repeatedly extracting the smallest element and adding the next element from the same list.

beginner

What data structure is commonly used to implement a min heap in C#?

A PriorityQueue is used in modern C# to implement a min heap, where elements are dequeued based on their priority (smallest priority first).

intermediate

In merging K sorted lists using a min heap, what do you store in the heap?

You store the current smallest elements from each list along with information about which list they came from, so you can add the next element from that list after extracting the smallest.

intermediate

Why is merging K sorted lists using a min heap more efficient than merging lists one by one?

Using a min heap reduces the time complexity to O(N log K), where N is total elements and K is number of lists, by always extracting the smallest element efficiently. Merging one by one can be slower because it repeatedly merges large lists.

beginner

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.

What does the min heap store when merging K sorted lists?

A All elements from all lists at once

B Only the largest elements from each list

C The smallest current elements from each list

D The middle elements of each list

The min heap stores the smallest current elements from each list to efficiently find the next smallest element.

What is the main benefit of using a min heap in merging K sorted lists?

A It reverses the lists

B It sorts all lists individually

C It merges lists without extra memory

D It keeps the smallest element accessible in O(1) time

A min heap keeps the smallest element at the top, allowing quick access to the smallest element.

What C# class is best suited to implement a min heap for this problem?

A Dictionary<TKey, TValue>

B PriorityQueue<TElement, TPriority>

C Stack<T>

D List<T>

PriorityQueue is designed to implement min heaps efficiently.

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

A O(N log K)

B O(K log N)

C O(N K)

D O(log N)

The time complexity is O(N log K), where N is total elements and K is number of lists.

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

A Add the next element from the same list to the min heap

B Add the largest element from any list

C Remove all elements from the heap

D Sort the heap again

After extracting the smallest element, add the next element from the same list to continue merging.

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

Think about how you pick the smallest number from multiple sorted lists.

Describe the steps to merge K sorted lists using a min heap in C#.

Imagine you have K friends each with sorted cards, and you pick the smallest card each time.

Practice

1. What is the main purpose of using a min heap when merging K sorted linked lists?

easy

A. To efficiently get the smallest current element among all lists

B. To store all elements in sorted order before merging

C. To reverse the order of elements in each list

D. To count the total number of elements in all lists

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 without scanning all elements.

2. Step 2: Why min heap is chosen

   It maintains the smallest element at the top, so we can pick and remove it efficiently while merging.

3. Final Answer:

   To efficiently get the smallest current element among all lists -> Option A

4. Quick Check:

   Min heap = smallest element retrieval [OK]

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

Common Mistakes:

• Thinking min heap stores all elements sorted
• Confusing min heap with max heap
• Using min heap to count elements

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

easy

A. var minHeap = new List<int>();

B. var minHeap = new SortedSet<(int val, int listIndex, ListNode node)>();

C. var minHeap = new Queue<ListNode>();

D. var minHeap = new Stack<ListNode>();

Solution

1. Step 1: Identify data structure for min heap

   SortedSet with tuple (value, list index, node) can act as a min heap by sorting elements by value.

2. Step 2: Why other options are incorrect

   List, Queue, and Stack do not maintain sorted order automatically, so they can't serve as min heap.

3. Final Answer:

   var minHeap = new SortedSet<(int val, int listIndex, ListNode node)>(); -> Option B

4. Quick Check:

   SortedSet = min heap structure [OK]

Hint: Use SortedSet with tuples for min heap in C# [OK]

Common Mistakes:

• Using List or Queue as min heap
• Not including value for sorting in tuple
• Confusing Stack with heap

3. What will be the output linked list after merging these 3 sorted lists using a min heap?

List 1: 1 -> 4 -> 5 -> null
List 2: 1 -> 3 -> 4 -> null
List 3: 2 -> 6 -> null

medium

A. 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null

B. 1 -> 3 -> 4 -> 1 -> 4 -> 5 -> 2 -> 6 -> null

C. 6 -> 5 -> 4 -> 4 -> 3 -> 2 -> 1 -> 1 -> null

D. 1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> null

Solution

1. Step 1: Insert first nodes of all lists into min heap

   Insert 1 (List1), 1 (List2), 2 (List3) into min heap.

2. Step 2: Extract min and add next nodes

   Extract 1 (List1), add 4; extract 1 (List2), add 3; extract 2 (List3), add 6; continue extracting min and adding next nodes until all lists are merged.

3. Final Answer:

   1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> null -> Option D

4. Quick Check:

   Merge sorted lists = sorted combined list [OK]

Hint: Always pick smallest node from heap and add next from same list [OK]

Common Mistakes:

• Not adding next node after extracting min
• Merging lists without sorting
• Confusing order of nodes

4. Given the following code snippet for merging K sorted lists, what is the error?

var minHeap = new SortedSet<(int val, int listIndex, ListNode node)>();
foreach (var list in lists) {
  if (list != null) minHeap.Add((list.val, 0, list));
}
while (minHeap.Count > 0) {
  var min = minHeap.Min;
  minHeap.Remove(min);
  // Add min.node to result
  if (min.node.next != null) {
    minHeap.Add((min.node.next.val, min.listIndex + 1, min.node.next));
  }
}

medium

A. Incrementing listIndex incorrectly when adding next node

B. Not checking if list is null before adding to heap

C. Using SortedSet instead of PriorityQueue

D. Not removing min element from heap

Solution

1. Step 1: Understand listIndex usage

   listIndex should identify which list the node belongs to and remain constant for that list.

2. Step 2: Identify error in incrementing listIndex

   Incrementing listIndex by 1 when adding next node is wrong; it should stay the same to track correct list.

3. Final Answer:

   Incrementing listIndex incorrectly when adding next node -> Option A

4. Quick Check:

   listIndex must not change per node [OK]

Hint: Keep listIndex fixed per list, don't increment per node [OK]

Common Mistakes:

• Changing listIndex for next nodes
• Forgetting to remove min element
• Ignoring null checks

5. You have 4 sorted linked lists with lengths 2, 3, 4, and 5 respectively. Using a min heap, what is the time complexity to merge all lists into one sorted list?

hard

A. O(N + K), where N is total nodes and K is number of lists

B. O(N K), where N is total nodes and K is number of lists

C. O(N log K), where N is total nodes and K is number of lists

D. O(K log N), where N is total nodes and K is number of lists

Solution

1. Step 1: Calculate total nodes and understand K

   Total nodes N = 2 + 3 + 4 + 5 = 14, K = 4 lists.

2. Step 2: Analyze min heap operations

   Each node is pushed and popped once from min heap of size K, so operations take O(log K) each.

3. Step 3: Combine to get total complexity

   Total time = O(N log K) because each of N nodes involves heap operations.

4. Final Answer:

   O(N log K), where N is total nodes and K is number of lists -> Option C

5. Quick Check:

   Heap operations per node = log K [OK]

Hint: Merging K lists with heap = O(N log K) time [OK]

Common Mistakes:

• Assuming O(N K) instead of O(N log K)
• Ignoring heap operation cost
• Confusing N and K roles