Merge K Sorted Lists Using Min Heap in DSA C++ - Build from Scratch

Merge K Sorted Lists Using Min Heap

📖 Scenario: You are working on a system that merges multiple sorted data streams into one sorted stream efficiently. This is common in search engines, data processing, and real-time analytics.

🎯 Goal: Build a program that merges k sorted linked lists into one sorted linked list using a min heap (priority queue) for efficient merging.

📋 What You'll Learn

Create k sorted linked lists with given values

Use a min heap (priority queue) to merge these lists efficiently

Output the merged sorted linked list

💡 Why This Matters

🌍 Real World

Merging multiple sorted data streams is common in search engines, log processing, and real-time data analytics where data arrives in sorted chunks.

💼 Career

Understanding how to merge sorted lists efficiently using heaps is important for software engineers working on databases, big data, and systems programming.

1

Create K Sorted Linked Lists

Create three sorted linked lists named list1, list2, and list3 with these exact values: - list1: 1 -> 4 -> 7 -> null - list2: 2 -> 5 -> 8 -> null - list3: 3 -> 6 -> 9 -> null Use the provided ListNode struct and create the lists by linking nodes manually.

DSA C++

#include 

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

int main() {
    // Create list1: 1 -> 4 -> 7 -> null
    // Create list2: 2 -> 5 -> 8 -> null
    // Create list3: 3 -> 6 -> 9 -> null
    // Your code here
    return 0;
}

Hint

Use the new keyword to create nodes and link them using the next pointer.

2

Store Lists in a Vector

Create a std::vector<ListNode*> named lists and add list1, list2, and list3 to it.

DSA C++

#include 
#include 

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

int main() {
    ListNode* list1 = new ListNode(1);
    list1->next = new ListNode(4);
    list1->next->next = new ListNode(7);

    ListNode* list2 = new ListNode(2);
    list2->next = new ListNode(5);
    list2->next->next = new ListNode(8);

    ListNode* list3 = new ListNode(3);
    list3->next = new ListNode(6);
    list3->next->next = new ListNode(9);

    // Create vector lists and add list1, list2, list3
    // Your code here
    return 0;
}

Hint

Use curly braces {} to initialize the vector with the three list pointers.

3

Merge Lists Using Min Heap

Write a function ListNode* mergeKLists(std::vector<ListNode*>& lists) that merges the k sorted linked lists using a std::priority_queue as a min heap. Use a custom comparator to order nodes by their val. Return the head of the merged sorted linked list.

DSA C++

#include 
#include 
#include 

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

std::vector lists = {};

ListNode* mergeKLists(std::vector& lists) {
    // Implement merge using min heap
    // Your code here
}

int main() {
    ListNode* list1 = new ListNode(1);
    list1->next = new ListNode(4);
    list1->next->next = new ListNode(7);

    ListNode* list2 = new ListNode(2);
    list2->next = new ListNode(5);
    list2->next->next = new ListNode(8);

    ListNode* list3 = new ListNode(3);
    list3->next = new ListNode(6);
    list3->next->next = new ListNode(9);

    std::vector lists = {list1, list2, list3};

    // Call mergeKLists here
    // Your code here
    return 0;
}

Hint

Use a priority queue with a custom comparator that compares ListNode values to always get the smallest node.

4

Print the Merged Sorted List

Write a while loop to print the merged linked list starting from mergedHead. Print each node's val followed by ->. End with null.

DSA C++

#include 
#include 
#include 

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

struct Compare {
    bool operator()(ListNode* a, ListNode* b) {
        return a->val > b->val;
    }
};

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

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

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

    while (!minHeap.empty()) {
        ListNode* smallest = minHeap.top();
        minHeap.pop();
        tail->next = smallest;
        tail = tail->next;
        if (smallest->next != nullptr) {
            minHeap.push(smallest->next);
        }
    }

    return dummy.next;
}

int main() {
    ListNode* list1 = new ListNode(1);
    list1->next = new ListNode(4);
    list1->next->next = new ListNode(7);

    ListNode* list2 = new ListNode(2);
    list2->next = new ListNode(5);
    list2->next->next = new ListNode(8);

    ListNode* list3 = new ListNode(3);
    list3->next = new ListNode(6);
    list3->next->next = new ListNode(9);

    std::vector lists = {list1, list2, list3};

    ListNode* mergedHead = mergeKLists(lists);

    // Print merged list
    // Your code here
    return 0;
}

Hint

Use a while loop to traverse the merged list and print each node's value followed by ->. End with null.

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