Merge K Sorted Lists Using Min Heap in DSA C++

Mental Model

We combine many sorted lists into one sorted list by always picking the smallest current element from all lists using a small helper structure.

Analogy: Imagine you have K friends each with a sorted list of candies by sweetness. You want to pick candies one by one in order of sweetness. You ask each friend for their smallest candy and pick the sweetest among those smallest candies, then ask that friend for their next candy, repeating until all candies are picked.

List1: 1 -> 4 -> 7 -> null
List2: 2 -> 5 -> 8 -> null
List3: 3 -> 6 -> 9 -> null

MinHeap: [1, 2, 3]

Merged: null

Dry Run Walkthrough

Input: lists: [1->4->7, 2->5->8, 3->6->9]

Goal: Merge all 3 sorted lists into one sorted list: 1->2->3->4->5->6->7->8->9

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

MinHeap: [1, 2, 3]
Merged: null

Why: We start by knowing the smallest elements from each list to pick the smallest overall

Step 2: Extract min (1) from heap and add to merged list; insert next node (4) from list1 into heap

MinHeap: [2, 3, 4]
Merged: 1 -> null

Why: We pick the smallest element and then add the next candidate from that list

Step 3: Extract min (2) from heap and add to merged list; insert next node (5) from list2 into heap

MinHeap: [3, 4, 5]
Merged: 1 -> 2 -> null

Why: Continue picking smallest and replenishing heap with next nodes

Step 4: Extract min (3) from heap and add to merged list; insert next node (6) from list3 into heap

MinHeap: [4, 5, 6]
Merged: 1 -> 2 -> 3 -> null

Why: Keep merging by always picking the smallest current element

Step 5: Extract min (4) from heap and add to merged list; insert next node (7) from list1 into heap

MinHeap: [5, 6, 7]
Merged: 1 -> 2 -> 3 -> 4 -> null

Why: Add next smallest and continue until all lists are exhausted

Step 6: Extract min (5) from heap and add to merged list; insert next node (8) from list2 into heap

MinHeap: [6, 7, 8]
Merged: 1 -> 2 -> 3 -> 4 -> 5 -> null

Why: Keep merging by replenishing heap with next nodes

Step 7: Extract min (6) from heap and add to merged list; insert next node (9) from list3 into heap

MinHeap: [7, 8, 9]
Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null

Why: Continue until all nodes are merged

Step 8: Extract min (7) from heap and add to merged list; list1 exhausted, no insertion

MinHeap: [8, 9]
Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null

Why: No more nodes in list1, just add remaining from others

Step 9: Extract min (8) from heap and add to merged list; list2 exhausted, no insertion

MinHeap: [9]
Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null

Why: Continue adding remaining nodes

Step 10: Extract min (9) from heap and add to merged list; list3 exhausted, heap empty

MinHeap: []
Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

Why: All nodes merged, heap empty, done

Result:

1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

Annotated Code

DSA C++

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

// Comparator for min heap to compare ListNode pointers by their val
struct Compare {
    bool operator()(ListNode* a, ListNode* b) {
        return a->val > b->val;
    }
};

ListNode* mergeKLists(vector<ListNode*>& lists) {
    priority_queue<ListNode*, vector<ListNode*>, Compare> minHeap;

    // Push first node of each list into min heap
    for (auto node : lists) {
        if (node != nullptr) {
            minHeap.push(node); // add smallest candidates
        }
    }

    ListNode dummy(0);
    ListNode* tail = &dummy;

    while (!minHeap.empty()) {
        ListNode* smallest = minHeap.top();
        minHeap.pop(); // get smallest node

        tail->next = smallest; // add to merged list
        tail = tail->next;

        if (smallest->next != nullptr) {
            minHeap.push(smallest->next); // add next node from same list
        }
    }

    return dummy.next;
}

void printList(ListNode* head) {
    while (head != nullptr) {
        cout << head->val;
        if (head->next != nullptr) cout << " -> ";
        head = head->next;
    }
    cout << " -> null" << endl;
}

int main() {
    // Create lists: [1->4->7], [2->5->8], [3->6->9]
    ListNode* l1 = new ListNode(1);
    l1->next = new ListNode(4);
    l1->next->next = new ListNode(7);

    ListNode* l2 = new ListNode(2);
    l2->next = new ListNode(5);
    l2->next->next = new ListNode(8);

    ListNode* l3 = new ListNode(3);
    l3->next = new ListNode(6);
    l3->next->next = new ListNode(9);

    vector<ListNode*> lists = {l1, l2, l3};

    ListNode* merged = mergeKLists(lists);
    printList(merged);

    return 0;
}

for (auto node : lists) { if (node != nullptr) { minHeap.push(node); } }

Add first node of each list to min heap to start merging

while (!minHeap.empty()) {

Repeat until all nodes are merged

ListNode* smallest = minHeap.top(); minHeap.pop();

Extract smallest node from min heap

tail->next = smallest; tail = tail->next;

Add smallest node to merged list and move tail pointer

if (smallest->next != nullptr) { minHeap.push(smallest->next); }

Add next node from the same list to min heap if exists

Output

1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

Complexity Analysis

Time: O(N log K) because we process all N nodes and each insertion/extraction from min heap of size K costs O(log K)

Space: O(K) because the min heap stores at most one node from each of the K lists at a time

vs Alternative: Naive merging by repeatedly merging two lists costs O(KN) which is slower; min heap approach is more efficient for large K

Edge Cases

Empty input lists vector

Returns null immediately, no nodes to merge

DSA C++

if (node != nullptr) { minHeap.push(node); }

Some lists are empty (null)

Only non-empty lists contribute nodes; empty lists ignored

DSA C++

if (node != nullptr) { minHeap.push(node); }

All lists empty

Returns null, merged list is empty

DSA C++

while (!minHeap.empty()) { ... }

When to Use This Pattern

When you need to merge multiple sorted lists efficiently, especially when K is large, use a min heap to always pick the smallest current element from all lists.

Common Mistakes

Mistake: Not pushing the next node of the extracted node back into the min heap
Fix: After extracting the smallest node, always push its next node into the min heap if it exists

Mistake: Using a max heap or incorrect comparator causing wrong order
Fix: Use a min heap with comparator that compares nodes by their value in ascending order

Summary

Merges K sorted linked lists into one sorted list using a min heap to efficiently pick the smallest node.

Use when you have multiple sorted lists and want to merge them quickly without repeatedly merging pairs.

The key insight is to keep track of the smallest current nodes from all lists in a min heap and always pick the smallest to build the merged list.

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 the heads of all lists.

B. It sorts each list individually before merging.

C. It uses extra memory 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 quickly find the smallest element among multiple sorted lists by always keeping the smallest element at the top.

2. Step 2: Compare with other options

   Options A, B, and C do not describe the min heap's advantage correctly. The min heap does not sort lists individually or concatenate without sorting.

3. Final Answer:

   It efficiently finds the smallest element among the heads of all lists. -> 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
• Assuming min heap stores all elements at once
• Confusing min heap with simple concatenation

2. Which of the following is the correct way to push an element into a min heap in C++ using STL?

easy

A. std::priority_queue minHeap; minHeap.push(5);

B. std::priority_queue<int> minHeap; minHeap.push(5);

C. std::priority_queue minHeap; minHeap.push(5);

D. std::priority_queue minHeap; minHeap.push(5);

Solution

1. Step 1: Identify min heap declaration syntax

   In C++, to create a min heap, use std::priority_queue with std::greater comparator and std::vector as container.

2. Step 2: Check each option

   std::priority_queue minHeap; minHeap.push(5); uses std::greater and vector correctly. std::priority_queue<int> minHeap; minHeap.push(5); is max heap by default. std::priority_queue minHeap; minHeap.push(5); uses std::less which is max heap. std::priority_queue minHeap; minHeap.push(5); uses std::list which is invalid container for priority_queue.

3. Final Answer:

   std::priority_queue minHeap; minHeap.push(5); -> Option C

4. Quick Check:

   Min heap = priority_queue + greater + vector [OK]

Hint: Min heap needs std::greater and vector container [OK]

Common Mistakes:

• Using default priority_queue (max heap)
• Using std::less instead of std::greater
• Using invalid container like std::list

3. What will be the output of the following code snippet that merges 3 sorted lists using a min heap?

#include <iostream>
#include <vector>
#include <queue>

struct Node {
    int val;
    int listIndex;
    int elementIndex;
    Node(int v, int li, int ei) : val(v), listIndex(li), elementIndex(ei) {}
};

struct Compare {
    bool operator()(Node const& a, Node const& b) {
        return a.val > b.val;
    }
};

int main() {
    std::vector lists = {{1,4,5}, {1,3,4}, {2,6}};
    std::priority_queue minHeap;

    for (int i = 0; i < lists.size(); i++) {
        if (!lists[i].empty())
            minHeap.push(Node(lists[i][0], i, 0));
    }

    std::vector<int> merged;
    while (!minHeap.empty()) {
        Node curr = minHeap.top();
        minHeap.pop();
        merged.push_back(curr.val);
        int nextIndex = curr.elementIndex + 1;
        if (nextIndex < lists[curr.listIndex].size()) {
            minHeap.push(Node(lists[curr.listIndex][nextIndex], curr.listIndex, nextIndex));
        }
    }

    for (int val : merged) std::cout << val << " ";
    return 0;
}

medium

A. 1 1 2 3 4 4 5 6

B. 1 2 3 4 4 5 6

C. 1 1 3 4 4 5 6 2

D. 2 1 1 3 4 4 5 6

Solution

1. Step 1: Understand the merging process

   The code pushes the first element of each list into the min heap. Then it repeatedly extracts the smallest element and pushes the next element from the same list until all elements are merged.

2. Step 2: Trace the merged output

   Lists: {1,4,5}, {1,3,4}, {2,6}. Merged order: 1 (from list0), 1 (list1), 2 (list2), 3 (list1), 4 (list0), 4 (list1), 5 (list0), 6 (list2).

3. Final Answer:

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

4. Quick Check:

   Merge sorted lists = 1 1 2 3 4 4 5 6 [OK]

Hint: Push first elements, pop min, push next from same list [OK]

Common Mistakes:

• Mixing order of elements
• Forgetting to push next element after pop
• Ignoring duplicates in merged output

4. Identify the bug in the following code snippet that attempts to merge K sorted lists using a min heap:

std::priority_queue minHeap;
for (auto& list : lists) {
    for (int val : list) {
        minHeap.push(val);
    }
}
std::vector<int> merged;
while (!minHeap.empty()) {
    merged.push_back(minHeap.top());
    minHeap.pop();
}

medium

A. The min heap is declared incorrectly and will cause a compilation error.

B. The code pushes all elements at once, losing the benefit of merging sorted lists efficiently.

C. The code does not pop elements from the min heap, causing an infinite loop.

D. The merged vector is not initialized before use.

Solution

1. Step 1: Analyze the approach used

   The code pushes all elements from all lists into the min heap at once, then pops them to get sorted order.

2. Step 2: Identify inefficiency

   This approach loses the efficiency of merging K sorted lists using min heap by only pushing the first element of each list and pushing next elements on demand.

3. Final Answer:

   The code pushes all elements at once, losing the benefit of merging sorted lists efficiently. -> Option B

4. Quick Check:

   Push all elements at once = inefficient merge [OK]

Hint: Push only first elements, then next after popping [OK]

Common Mistakes:

• Pushing all elements at once
• Ignoring heap declaration correctness
• Forgetting to pop elements in loop

5. Given K sorted linked lists, how can you modify the min heap approach to merge them efficiently if some lists are empty and others have duplicate values?

hard

A. Ignore empty lists and merge only non-empty lists by concatenation without using a min heap.

B. Use a max heap instead of min heap to handle duplicates and empty lists.

C. Push all elements from all lists into the min heap at once, then pop to merge, skipping duplicates manually.

D. Initialize the min heap with the first non-empty element of each list and push next elements only when popping, allowing duplicates naturally.

Solution

1. Step 1: Handle empty lists and duplicates

   Skip empty lists when initializing the min heap. For duplicates, min heap naturally handles them by ordering values.

2. Step 2: Efficient merging strategy

   Push only the first element of each non-empty list into the min heap. When popping, push the next element from the same list if it exists. This keeps memory and operations efficient.

3. Final Answer:

   Initialize the min heap with the first non-empty element of each list and push next elements only when popping, allowing duplicates naturally. -> Option D

4. Quick Check:

   Push first non-empty elements, push next on pop, duplicates allowed [OK]

Hint: Skip empty lists, push next element after pop [OK]

Common Mistakes:

• Pushing all elements at once wasting memory
• Ignoring empty lists causing errors
• Using max heap which breaks order