Merge K Sorted Lists Using Min Heap in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind using a min heap to merge K sorted lists?

A min heap helps efficiently find the smallest element among the heads of all lists, allowing us to build the merged sorted list by always picking the smallest next element.

beginner

In the context of merging K sorted lists, what does each element in the min heap represent?

Each element in the min heap represents the current node (or value) from one of the K lists that is a candidate to be added next to the merged list.

intermediate

Why is the time complexity of merging K sorted lists using a min heap O(N log K)?

Because we insert and remove each of the N total elements once from a min heap of size at most K, and each heap operation takes O(log K) time.

beginner

What happens after extracting the smallest element from the min heap during the merge process?

After extracting the smallest element, we add it to the merged list and then insert the next element from the same list (if any) into the min heap.

beginner

How does the min heap help maintain the sorted order in the merged list?

The min heap always keeps the smallest current elements at the top, so extracting from it gives the next smallest element to add, preserving sorted order.

What data structure is best suited to efficiently merge K sorted lists?

A Queue

B Stack

C Hash map

D Min heap

A min heap efficiently finds the smallest element among multiple lists, making it ideal for merging sorted lists.

What is the maximum size of the min heap during the merge of K sorted lists?

A N (total elements)

B K (number of lists)

C 1

D N/K

The min heap holds at most one element from each of the K lists at any time.

After extracting the smallest element from the min heap, what is the next step?

A Insert the next element from the same list into the min heap

B Insert the largest element from all lists

C Clear the min heap

D Stop the process

We insert the next element from the same list to continue merging in sorted order.

What is the overall time complexity of merging K sorted lists with total N elements using a min heap?

A O(N log N)

B O(K log N)

C O(N log K)

D O(N + K)

Each of the N elements is pushed and popped from a min heap of size K, costing O(log K) per operation.

Why is a min heap preferred over a simple linear search to find the smallest element among K lists?

A Min heap reduces the search time to O(log K) per operation

B Linear search is faster

C Min heap uses more memory

D Linear search is easier to implement

Min heap allows efficient retrieval of the smallest element in O(log K) time, better than O(K) for linear search.

Explain how a min heap is used to merge K sorted lists step-by-step.

Think about how the min heap always gives the smallest current element.

Describe the time complexity of merging K sorted lists using a min heap and why it is efficient.

Focus on how many times heap operations happen and their cost.

Practice

1. What is the main advantage of using a min heap to merge K sorted lists?

easy

A. It uses extra memory to store all elements at once.

B. It sorts each list individually before merging.

C. It efficiently finds the smallest element among the heads of all lists.

D. It merges lists by concatenating them without sorting.

Solution

1. Step 1: Understand the role of a min heap in merging

   A min heap keeps track of the smallest elements efficiently, allowing quick access to the smallest head among K lists.

2. Step 2: Compare other options

   Options B, C, and D do not describe the min heap's role correctly. Sorting individually or concatenating without sorting does not merge sorted lists efficiently.

3. Final Answer:

   It efficiently finds the smallest element among the heads of all lists. -> Option C

4. Quick Check:

   Min heap = efficient smallest element access [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 Go data structure is best suited to implement a min heap for merging K sorted lists?

easy

A. map

B. linked list

C. array without heap functions

D. slice with container/heap package

Solution

1. Step 1: Identify Go's min heap implementation

   Go uses slices combined with the container/heap package to implement min heaps efficiently.

2. Step 2: Evaluate other options

   Maps and linked lists do not provide heap properties. Arrays without heap functions lack automatic heap behavior.

3. Final Answer:

   slice with container/heap package -> Option D

4. Quick Check:

   Go min heap = slice + container/heap [OK]

Hint: Use container/heap with slice for min heap in Go [OK]

Common Mistakes:

• Using map which has no order
• Using linked list which is slow for heap ops
• Ignoring container/heap package

3. What is the output of merging these 3 sorted lists using a min heap?

list1 = [1, 4, 7]
list2 = [2, 5, 8]
list3 = [3, 6, 9]

medium

A. [1, 3, 5, 7, 9, 2, 4, 6, 8]

B. [1, 2, 3, 4, 5, 6, 7, 8, 9]

C. [9, 8, 7, 6, 5, 4, 3, 2, 1]

D. [1, 4, 7, 2, 5, 8, 3, 6, 9]

Solution

1. Step 1: Understand merging with min heap

   The min heap picks the smallest head element from all lists repeatedly, producing a fully sorted merged list.

2. Step 2: Merge given lists step-by-step

   Starting with heads: 1, 2, 3; pick 1, then 2, then 3, then 4, 5, 6, 7, 8, 9 in order.

3. Final Answer:

   [1, 2, 3, 4, 5, 6, 7, 8, 9] -> Option B

4. Quick Check:

   Merge sorted lists = sorted combined list [OK]

Hint: Min heap merges by always picking smallest next element [OK]

Common Mistakes:

• Concatenating lists without sorting
• Reversing order incorrectly
• Mixing elements from different lists randomly

4. Given this Go snippet for merging K sorted lists, what is the bug?

func mergeKLists(lists [][]int) []int {
    var result []int
    h := &IntHeap{}
    heap.Init(h)
    for _, list := range lists {
        for _, val := range list {
            heap.Push(h, val)
        }
    }
    for h.Len() > 0 {
        result = append(result, heap.Pop(h).(int))
    }
    return result
}

medium

A. Pushing all elements at once loses the benefit of merging sorted lists.

B. Heap is not initialized before pushing elements.

C. Result slice is not declared properly.

D. Heap.Pop returns wrong type.

Solution

1. Step 1: Analyze the merging approach

   The code pushes all elements from all lists into the heap at once, ignoring the sorted order advantage.

2. Step 2: Understand the impact

   This approach works but is inefficient (O(N log N)) instead of O(N log K) by pushing only heads and updating heap dynamically.

3. Final Answer:

   Pushing all elements at once loses the benefit of merging sorted lists. -> Option A

4. Quick Check:

   Push all elements = inefficient merge [OK]

Hint: Push only heads, not all elements at once [OK]

Common Mistakes:

• Ignoring heap.Init call
• Misunderstanding result slice declaration
• Incorrect type assertion on Pop

5. You have K sorted linked lists of varying lengths. How would you efficiently merge them using a min heap in Go to minimize time complexity?

hard

A. Push the first node of each list into the min heap, then repeatedly pop the smallest and push its next node.

B. Push all nodes from all lists into the min heap at once, then pop all to get merged list.

C. Concatenate all lists and sort the combined list after.

D. Use a max heap instead of a min heap for merging.

Solution

1. Step 1: Use min heap with heads of each list

   Push only the first node of each list into the min heap to keep track of the smallest current element among lists.

2. Step 2: Pop smallest and push next node

   Pop the smallest node, add it to the merged list, then push the next node from that list into the heap, repeating until all nodes are processed.

3. Step 3: Time complexity advantage

   This approach runs in O(N log K) time, where N is total nodes and K is number of lists, which is efficient.

4. Final Answer:

   Push the first node of each list into the min heap, then repeatedly pop the smallest and push its next node. -> Option A

5. Quick Check:

   Push heads + pop/push next = efficient merge [OK]

Hint: Push heads only, then push next nodes dynamically [OK]

Common Mistakes:

• Pushing all nodes at once causing inefficiency
• Using max heap which gives largest element first
• Sorting concatenated list which is slower