Merge K Sorted Lists Using Min Heap in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of merging 3 sorted lists using min heap

What is the printed merged list after running the following Go code that merges 3 sorted linked lists using a min heap?

DSA Go

package main

import (
	"container/heap"
	"fmt"
)

type ListNode struct {
	Val  int
	Next *ListNode
}

type MinHeap []*ListNode

func (h MinHeap) Len() int           { return len(h) }
func (h MinHeap) Less(i, j int) bool { return h[i].Val < h[j].Val }
func (h MinHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *MinHeap) Push(x interface{}) {
	*h = append(*h, x.(*ListNode))
}

func (h *MinHeap) Pop() interface{} {
	n := len(*h)
	item := (*h)[n-1]
	*h = (*h)[:n-1]
	return item
}

func mergeKLists(lists []*ListNode) *ListNode {
	h := &MinHeap{}
	heap.Init(h)

	for _, node := range lists {
		if node != nil {
			heap.Push(h, node)
		}
	}

	dummy := &ListNode{}
	current := dummy

	for h.Len() > 0 {
		minNode := heap.Pop(h).(*ListNode)
		current.Next = minNode
		current = current.Next
		if minNode.Next != nil {
			heap.Push(h, minNode.Next)
		}
	}

	return dummy.Next
}

func printList(head *ListNode) {
	for head != nil {
		fmt.Printf("%d -> ", head.Val)
		head = head.Next
	}
	fmt.Println("null")
}

func main() {
	list1 := &ListNode{1, &ListNode{4, &ListNode{5, nil}}}
	list2 := &ListNode{1, &ListNode{3, &ListNode{4, nil}}}
	list3 := &ListNode{2, &ListNode{6, nil}}

	merged := mergeKLists([]*ListNode{list1, list2, list3})
	printList(merged)
}

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

B 1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> null

C 1 -> 1 -> 3 -> 4 -> 4 -> 5 -> 6 -> null

D 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> null

💡 Hint

Remember the min heap always pops the smallest current node from all lists.

The min heap merges nodes by always choosing the smallest current node among all lists. The merged list contains all nodes in sorted order: 1,1,2,3,4,4,5,6.

🧠 Conceptual

intermediate

Understanding the role of the min heap in merging K sorted lists

Why is a min heap used to merge K sorted linked lists efficiently?

A Because it merges lists by concatenating them directly

B Because it sorts all elements at once in O(N log N) time

C Because it stores all elements in a balanced binary search tree

D Because it always provides the smallest current element among all lists in O(log K) time

💡 Hint

Think about how to get the smallest element quickly from multiple sorted lists.

A min heap keeps track of the smallest current element from each list, allowing extraction in O(log K) time, making the overall merge efficient.

🔧 Debug

advanced

Identify the error in this min heap implementation for merging lists

What error will occur when running this Go code snippet for merging K sorted lists using a min heap?

DSA Go

type MinHeap []*ListNode

func (h MinHeap) Len() int           { return len(h) }
func (h MinHeap) Less(i, j int) bool { return h[i].Val > h[j].Val }
func (h MinHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

A The merged list will be sorted in descending order, not ascending

B Compilation error due to incorrect method signatures

C Runtime panic due to nil pointer dereference

D The heap will not compile because Less method returns bool incorrectly

💡 Hint

Check the Less method logic and how it affects heap order.

The Less method uses '>' which makes the heap a max heap, so the smallest element is not popped first, resulting in descending order merge.

🚀 Application

advanced

How many nodes are in the merged list after merging these K sorted lists?

Given 4 sorted linked lists with lengths 3, 2, 4, and 1 respectively, how many nodes will the merged list contain after merging using a min heap?

A 10

B 9

C 8

D 11

💡 Hint

Add the lengths of all lists to find total nodes.

The merged list contains all nodes from all lists combined, so total nodes = 3 + 2 + 4 + 1 = 10.

❓ Predict Output

expert

Output of merging empty and non-empty lists using min heap

What is the output of the following Go code that merges 3 lists where one is empty and others have nodes?

DSA Go

package main

import (
	"container/heap"
	"fmt"
)

type ListNode struct {
	Val  int
	Next *ListNode
}

type MinHeap []*ListNode

func (h MinHeap) Len() int           { return len(h) }
func (h MinHeap) Less(i, j int) bool { return h[i].Val < h[j].Val }
func (h MinHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *MinHeap) Push(x interface{}) {
	*h = append(*h, x.(*ListNode))
}

func (h *MinHeap) Pop() interface{} {
	n := len(*h)
	item := (*h)[n-1]
	*h = (*h)[:n-1]
	return item
}

func mergeKLists(lists []*ListNode) *ListNode {
	h := &MinHeap{}
	heap.Init(h)

	for _, node := range lists {
		if node != nil {
			heap.Push(h, node)
		}
	}

	dummy := &ListNode{}
	current := dummy

	for h.Len() > 0 {
		minNode := heap.Pop(h).(*ListNode)
		current.Next = minNode
		current = current.Next
		if minNode.Next != nil {
			heap.Push(h, minNode.Next)
		}
	}

	return dummy.Next
}

func printList(head *ListNode) {
	for head != nil {
		fmt.Printf("%d -> ", head.Val)
		head = head.Next
	}
	fmt.Println("null")
}

func main() {
	list1 := (*ListNode)(nil)
	list2 := &ListNode{0, &ListNode{2, nil}}
	list3 := &ListNode{1, nil}

	merged := mergeKLists([]*ListNode{list1, list2, list3})
	printList(merged)
}

A 0 -> 2 -> 1 -> null

B 1 -> 0 -> 2 -> null

C 0 -> 1 -> 2 -> null

D null

💡 Hint

Check how nil lists are skipped and how min heap orders nodes.

The merged list contains nodes 0, 1, and 2 in ascending order: 0 -> 1 -> 2 -> null.

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