Merge K Sorted Lists Using Min Heap in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of merging 3 sorted lists using min heap

What is 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

DSA C

/* Pseudocode for merging 3 sorted lists using min heap */
// Insert first node of each list into min heap
// Extract min node, add to result, insert next node from same list
// Repeat until heap empty

// Final merged list output:

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

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

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

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

💡 Hint

Remember to include all nodes from all lists in sorted order.

Merging 3 sorted lists using a min heap results in a fully sorted list containing all nodes from the input lists. The min heap always picks the smallest current node among the lists.

🧠 Conceptual

intermediate

Why use a min heap to merge K sorted lists?

Which of the following best explains why a min heap is used to merge K sorted lists efficiently?

A It stores all nodes in memory to avoid repeated traversal of lists.

B It sorts all nodes at once before merging, which is faster than merging step-by-step.

C It allows quick access to the smallest current node among all lists, reducing overall comparisons.

D It merges lists by comparing only the first and last nodes of each list.

💡 Hint

Think about how to always pick the smallest next node efficiently.

A min heap keeps track of the smallest node among the heads of all lists, allowing efficient extraction and insertion in O(log K) time, which reduces total comparisons.

🔧 Debug

advanced

Identify the error in this min heap merge code snippet

What error will this C code snippet cause when merging K sorted lists using a min heap? ```c struct Node* mergeKLists(struct Node* lists[], int k) { struct MinHeap* heap = createMinHeap(k); for (int i = 0; i < k; i++) { if (lists[i] != NULL) { insertMinHeap(heap, lists[i]); } } struct Node* head = NULL, *tail = NULL; while (!isEmpty(heap)) { struct Node* minNode = extractMin(heap); if (head == NULL) { head = tail = minNode; } else { tail->next = minNode; tail = minNode; } if (minNode->next != NULL) { insertMinHeap(heap, minNode->next); } } return head; } ```

A The merged list may form a cycle because next pointers are not set to NULL after extraction.

B The tail pointer is not updated correctly causing a memory leak.

C The code will skip some nodes because it does not insert all nodes into the heap.

D The min heap is not initialized properly causing a segmentation fault.

💡 Hint

Check what happens to the next pointers of extracted nodes.

Since the extracted minNode's next pointer is not set to NULL before linking, the merged list may contain cycles or incorrect links.

❓ Predict Output

advanced

Output after merging 4 sorted lists with duplicates

Given these 4 sorted lists, what is the output after merging using a min heap?

List 1: 2 -> 2 -> 5 -> null
List 2: 1 -> 3 -> 4 -> null
List 3: 2 -> 6 -> 7 -> null
List 4: 0 -> 8 -> null

A 1 -> 2 -> 2 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null

B 0 -> 1 -> 2 -> 2 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null

C 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null

D 0 -> 1 -> 2 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null

💡 Hint

Include all duplicates from all lists in sorted order.

All nodes including duplicates are merged in sorted order using the min heap, so all three '2's appear in the final list.

🧠 Conceptual

expert

Time complexity of merging K sorted lists using a min heap

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

A O(N + K log N)

B O(K log N)

C O(N K)

D O(N log K)

💡 Hint

Consider how many times nodes are inserted and extracted from the heap and the heap size.

Each of the N nodes is inserted and extracted once from a min heap of size at most K, so operations take O(log K) each, total O(N log K).

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