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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push the first element of each list into the min heap.

DSA C++

for (int i = 0; i < lists.size(); i++) {
    if (lists[i] != nullptr) {
        minHeap.[1](lists[i]);
    }
}

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

Aempty

Bpop

Ctop

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push to add elements.

Using top which only accesses the smallest element without adding.

We use push() to add elements to the min heap.

2fill in blank

medium

Complete the code to get the smallest node from the min heap.

DSA C++

while (!minHeap.empty()) {
    ListNode* node = minHeap.[1]();
    minHeap.pop();
    // process node
}

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

Atop

Bpop

Cpush

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() to get the element, which removes it immediately.

Using push() which adds elements instead of accessing.

top() returns the smallest element in the min heap without removing it.

3fill in blank

hard

Fix the error in the comparator struct for the min heap.

DSA C++

struct compare {
    bool operator() (ListNode* a, ListNode* b) {
        return a->val [1] b->val;
    }
};

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

A==

B>

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' which creates a max heap instead of min heap.

Using '==' or '!=' which do not define order.

For a min heap, the comparator returns true if a's value is greater than b's value.

4fill in blank

hard

Fill both blanks to correctly add the next node to the min heap after popping the smallest node.

DSA C++

if (node->[1] != nullptr) {
    minHeap.[2](node->next);
}

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

Anext

Bprev

Cpush

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' to access the next node.

Using pop instead of push to add the node.

We check if node->next exists and then push it into the min heap.

5fill in blank

hard

Fill all three blanks to correctly build the merged linked list.

DSA C++

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

while (!minHeap.empty()) {
    ListNode* node = minHeap.[1]();
    minHeap.pop();
    tail->[2] = node;
    tail = tail->[3];

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

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

Atop

Bnext

Cval

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() instead of top() to get the node.

Using 'val' or 'prev' instead of 'next' for linking.

We get the smallest node with top(), link it by setting tail->next, and move tail to tail->next.

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