Merge K Sorted Lists Using Min Heap in DSA C - Time & Space Complexity

Time Complexity: Merge K Sorted Lists Using Min Heap

O(N log K)

Understanding Time Complexity

We want to know how the time needed to merge multiple sorted lists grows as the number of lists and their sizes increase.

Specifically, how does using a min heap affect the speed of merging K sorted lists?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Merge K sorted linked lists using a min heap
struct ListNode* mergeKLists(struct ListNode** lists, int k) {
    // Create a min heap
    MinHeap* heap = createMinHeap(k);
    for (int i = 0; i < k; i++) {
        if (lists[i] != NULL) {
            insertMinHeap(heap, lists[i]);
        }
    }
    struct ListNode dummy;
    struct ListNode* tail = &dummy;
    while (!isEmpty(heap)) {
        struct ListNode* minNode = extractMin(heap);
        tail->next = minNode;
        tail = tail->next;
        if (minNode->next != NULL) {
            insertMinHeap(heap, minNode->next);
        }
    }
    return dummy.next;
}
    

This code merges K sorted linked lists by always picking the smallest current node using a min heap.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Extracting the smallest node from the min heap and inserting the next node from that list.
• How many times: This happens once for each of the total N nodes across all lists.

How Execution Grows With Input

Each node is inserted and extracted from the heap once. The heap size is at most K.

Input Size (N total nodes)	Approx. Operations (heap insert/extract)
10 nodes, K=3 lists	About 10 insertions and 10 extractions, each costing log 3 ≈ 1.6 steps
100 nodes, K=5 lists	About 100 insertions and 100 extractions, each costing log 5 ≈ 2.3 steps
1000 nodes, K=10 lists	About 1000 insertions and 1000 extractions, each costing log 10 ≈ 3.3 steps

Pattern observation: The total work grows roughly proportional to N times log K, where N is total nodes and K is number of lists.

Final Time Complexity

Time Complexity: O(N log K)

This means the time grows with the total number of nodes times the logarithm of the number of lists, making it efficient for many lists.

Common Mistake

[X] Wrong: "Merging K lists is just K times merging two lists, so it's O(KN)."

[OK] Correct: Repeatedly merging two lists one by one leads to more work. Using a min heap merges all lists simultaneously, reducing repeated scanning and making it faster than simple repeated merges.

Interview Connect

Understanding this time complexity helps you explain why using a min heap is a smart choice for merging multiple sorted lists efficiently in real projects.

Self-Check

What if we used a simple array to find the minimum node instead of a min heap? How would the time complexity change?

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