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

Time Complexity: Merge K Sorted Lists Using Min Heap

O(n log k)

Understanding Time Complexity

We want to understand how long it takes to merge multiple sorted lists into one sorted list using a min heap.

Specifically, how the time grows as the number of lists and their sizes increase.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

import heapq

def merge_k_lists(lists):
    min_heap = []
    for i, lst in enumerate(lists):
        if lst:
            heapq.heappush(min_heap, (lst[0], i, 0))
    result = []
    while min_heap:
        val, list_idx, element_idx = heapq.heappop(min_heap)
        result.append(val)
        if element_idx + 1 < len(lists[list_idx]):
            next_val = lists[list_idx][element_idx + 1]
            heapq.heappush(min_heap, (next_val, list_idx, element_idx + 1))
    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 sum of all 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=2)	About 10 push/pop operations on heap of size 2
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 total operations grow linearly with total elements n, but each heap operation depends on k, which is usually smaller.

Final Time Complexity

Time Complexity: O(n log k)

This means the time grows mostly with the total number of elements n, and each operation is a quick log k step because the heap size depends on the number of lists 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 multiple sorted lists, a common problem in coding interviews and real-world data processing.

Self-Check

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

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 individual list before merging.

C. It uses extra space to store all elements at once.

D. It merges lists by concatenating them directly.

Solution

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

   A min heap helps keep track of the smallest element among multiple sorted lists efficiently.

2. Step 2: Compare with other options

   Options B, C, and D do not describe the min heap's main advantage in this context.

3. Final Answer:

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

4. Quick Check:

   Min heap = smallest element tracking [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 concatenation with merging

2. Which of the following Python code snippets correctly initializes a min heap with the first elements of each sorted list named lists?

easy

A. heap = [lst[0] for lst in lists if lst]; import heapq; heapq.heapify(heap)

B. heap = lists[0]; import heapq; heapq.heapify(heap)

C. heap = []; for lst in lists: heap.append(lst); import heapq; heapq.heapify(heap)

D. heap = lists; import heapq; heapq.heapify(heap)

Solution

1. Step 1: Extract first elements safely

   heap = [lst[0] for lst in lists if lst]; import heapq; heapq.heapify(heap) uses list comprehension to get first elements only if the list is not empty.

2. Step 2: Check heap initialization

   heap = [lst[0] for lst in lists if lst]; import heapq; heapq.heapify(heap) then calls heapq.heapify to create a min heap from these elements correctly.

3. Final Answer:

   heap = [lst[0] for lst in lists if lst]; import heapq; heapq.heapify(heap) -> Option A

4. Quick Check:

   Correct min heap initialization = heap = [lst[0] for lst in lists if lst]; import heapq; heapq.heapify(heap) [OK]

Hint: Use list comprehension with condition and heapq.heapify [OK]

Common Mistakes:

• Not checking if list is empty before accessing lst[0]
• Heapifying entire lists instead of first elements
• Assigning heap to a single list instead of elements

3. What will be the output of the following code snippet that merges 3 sorted lists using a min heap?

import heapq
lists = [[1,4,5],[1,3,4],[2,6]]
heap = []
for i, lst in enumerate(lists):
    if lst:
        heapq.heappush(heap, (lst[0], i, 0))
result = []
while heap:
    val, list_idx, element_idx = heapq.heappop(heap)
    result.append(val)
    if element_idx + 1 < len(lists[list_idx]):
        next_tuple = (lists[list_idx][element_idx+1], list_idx, element_idx+1)
        heapq.heappush(heap, next_tuple)
print(result)

medium

A. [1, 2, 3, 4, 5, 6]

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

C. [1, 1, 3, 4, 4, 5, 6]

D. [1, 1, 2, 3, 4, 4, 5, 6]

Solution

1. Step 1: Trace heap operations

   Initially, heap contains (1,0,0), (1,1,0), (2,2,0). We pop smallest 1, add next element 4 from list 0, then pop next 1, add 3 from list 1, and so on.

2. Step 2: Build merged list step-by-step

   Result builds as [1,1,2,3,4,4,5,6] by always popping smallest and pushing next from same list.

3. Final Answer:

   [1, 1, 2, 3, 4, 4, 5, 6] -> Option D

4. Quick Check:

   Merged sorted list = [1,1,2,3,4,4,5,6] [OK]

Hint: Track heap pops and pushes carefully [OK]

Common Mistakes:

• Missing duplicate 1s in output
• Forgetting to push next element from same list
• Adding elements out of order

4. Identify the error in the following code snippet that attempts to merge k sorted lists using a min heap:

import heapq
lists = [[1,3,5],[2,4,6],[0,7,8]]
heap = []
for lst in lists:
    heapq.heappush(heap, lst[0])
result = []
while heap:
    val = heapq.heappop(heap)
    result.append(val)
print(result)

medium

A. The lists variable is not defined.

B. The heap is not initialized using heapq.heapify().

C. The code pushes only first elements and never pushes next elements from lists.

D. The code uses heappop instead of pop.

Solution

1. Step 1: Analyze heap push operations

   The code pushes only the first element of each list into the heap once and never pushes subsequent elements.

2. Step 2: Understand why this causes incomplete merge

   Without pushing next elements, only first elements are merged, so the result is incomplete and incorrect.

3. Final Answer:

   The code pushes only first elements and never pushes next elements from lists. -> Option C

4. Quick Check:

   Missing next elements push = The code pushes only first elements and never pushes next elements from lists. [OK]

Hint: Always push next elements after popping from heap [OK]

Common Mistakes:

• Assuming heapq.heapify is mandatory here
• Confusing heappop with pop
• Ignoring the need to push next elements

5. Given k sorted linked lists, which approach best describes how to merge them efficiently using a min heap? A) Insert all nodes from all lists into a min heap, then pop all to form merged list. B) Push only the first node of each list into the min heap, then repeatedly pop the smallest node and push its next node until all nodes are merged. C) Merge lists two at a time using simple merge, then merge results. D) Concatenate all lists and sort the combined list.

hard

A. Insert all nodes at once into min heap and pop all.

B. Push first nodes, pop smallest, push next nodes repeatedly.

C. Merge lists two at a time without heap.

D. Concatenate all lists and sort at the end.

Solution

1. Step 1: Understand efficient min heap usage

   Pushing only the first node of each list keeps the heap small and efficient.

2. Step 2: Describe the merging process

   Pop smallest node, add it to merged list, then push the next node from that list to heap, repeating until done.

3. Step 3: Compare with other options

   Insert all nodes at once into min heap and pop all. wastes space and time by pushing all nodes at once. Options B and D do not use min heap.

4. Final Answer:

   Push first nodes, pop smallest, push next nodes repeatedly. -> Option B

5. Quick Check:

   Efficient merge = push first nodes, pop smallest, push next [OK]

Hint: Push first nodes only, then push next after popping [OK]

Common Mistakes:

• Pushing all nodes at once wastes memory
• Not pushing next nodes after popping
• Using simple merge instead of heap for k lists