Merge K Sorted Lists Using Min Heap in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the min heap size.

DSA C

int heapSize = [1];

Drag options to blanks, or click blank then click option'

A0

Bk

C1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing heap size to 0 or 1 instead of k.

Using n (total elements) instead of k.

The heap size should be initialized to k, the number of sorted lists to merge.

2fill in blank

medium

Complete the code to extract the minimum element from the heap.

DSA C

Node* minNode = [1](minHeap);

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AinsertHeap

Bheapify

CextractMin

DbuildHeap

Attempts:

3 left

💡 Hint

Common Mistakes

Using insertHeap instead of extractMin.

Confusing heapify with extractMin.

The function extractMin removes and returns the smallest element from the min heap.

3fill in blank

hard

Fix the error in the code to insert the next element from the extracted node's list into the heap.

DSA C

if (minNode->next != NULL) {
    [1](minHeap, minNode->next);
}

Drag options to blanks, or click blank then click option'

AextractMin

BbuildHeap

CdeleteNode

DinsertHeap

Attempts:

3 left

💡 Hint

Common Mistakes

Using extractMin instead of insertHeap.

Calling buildHeap unnecessarily.

After extracting the minimum node, we insert its next node into the heap using insertHeap.

4fill in blank

hard

Fill both blanks to correctly build the min heap and initialize the array of heap nodes.

DSA C

Node* heapNodes = malloc(sizeof(Node*) * [1]);
buildHeap(heapNodes, [2]);

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Ak

Bn

CheapSize

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n (total elements) instead of k for array size.

Passing wrong size to buildHeap.

The array heapNodes should have size k, and buildHeap is called with heapSize to organize the heap.

5fill in blank

hard

Fill all three blanks to correctly update the merged list and heap after extracting the minimum node.

DSA C

mergedTail->next = [1];
mergedTail = mergedTail->next;
if ([2] != NULL) {
    [3](minHeap, [2]);
}

Drag options to blanks, or click blank then click option'

AminNode

BminNode->next

CinsertHeap

DextractMin

Attempts:

3 left

💡 Hint

Common Mistakes

Using extractMin instead of insertHeap for insertion.

Not updating mergedTail correctly.

We link the extracted minNode to the merged list, move the tail, and if the next node exists, insert it into the heap using insertHeap.

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