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Merge K Sorted Lists Using Min Heap in DSA Go - Time & Space Complexity

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Time Complexity: Merge K Sorted Lists Using Min Heap
O(n log k)
Understanding Time Complexity

We want to know how long it takes to merge multiple sorted lists using a min heap.

Specifically, how does the time grow as the number of lists and their sizes increase?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


import "container/heap"

type Item struct {
    value       int
    listIndex   int
    elementIndex int
}

type IntHeap []*Item

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

func (h *IntHeap) Push(x interface{}) {
    *h = append(*h, x.(*Item))
}

func (h *IntHeap) Pop() interface{} {
    old := *h
    n := len(old)
    item := old[n-1]
    *h = old[0 : n-1]
    return item
}

func mergeKLists(lists [][]int) []int {
    minHeap := &IntHeap{}
    heap.Init(minHeap)
    for i, list := range lists {
        if len(list) > 0 {
            heap.Push(minHeap, &Item{value: list[0], listIndex: i, elementIndex: 0})
        }
    }
    var result []int
    for minHeap.Len() > 0 {
        item := heap.Pop(minHeap).(*Item)
        result = append(result, item.value)
        if item.elementIndex+1 < len(lists[item.listIndex]) {
            nextIndex := item.elementIndex + 1
            heap.Push(minHeap, &Item{value: lists[item.listIndex][nextIndex], listIndex: item.listIndex, elementIndex: nextIndex})
        }
    }
    return result
}

This code merges k sorted lists into one sorted list using a min heap to always pick the smallest next element.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Removing the smallest element from the heap and adding the next element from the same list.
  • How many times: This happens once for every element across all lists, so total n times where n is the total number of elements.
How Execution Grows With Input

Each element is pushed and popped from the heap once. The heap size is at most k (number of lists).

Input Size (n)Approx. Operations
10 (k=3)About 10 push/pop operations on heap of size ≤ 3
100 (k=5)About 100 push/pop operations on heap of size ≤ 5
1000 (k=10)About 1000 push/pop operations on heap of size ≤ 10

Pattern observation: The number of operations grows linearly with total elements, but each heap operation depends on k, which is usually much smaller than n.

Final Time Complexity

Time Complexity: O(n log k)

This means the time grows mostly with the total number of elements n, but each step involves a quick operation on a heap of size k.

Common Mistake

[X] Wrong: "The time complexity is O(n log n) because we use a heap for all elements."

[OK] Correct: The heap never holds all n elements at once, only up to k elements, so the log factor depends on k, not n.

Interview Connect

Understanding this time complexity helps you explain efficient merging of sorted data streams, a common real-world problem.

Self-Check

"What if we used a simple list instead of a min heap to find the smallest element each time? How would the time complexity change?"

Practice

(1/5)
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
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