Merge K Sorted Lists Using Min Heap in DSA Python - 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 Python

for i in range(len(lists)):
    if lists[i]:
        heapq.heappush([1], (lists[i][0], i, 0))

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

Aresult

Blists

Cqueue

Dmin_heap

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable name instead of the heap.

Pushing the entire list instead of the first element.

We use 'min_heap' to store the first element of each list along with its list index and element index.

2fill in blank

medium

Complete the code to pop the smallest element from the min heap.

DSA Python

while min_heap:
    val, list_idx, element_idx = heapq.[1](min_heap)

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

Aheappop

Breplace

Cpeek

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'push' instead of 'heappop'.

Using 'peek' which does not remove the element.

We use 'heappop' to remove and return the smallest element from the min heap.

3fill in blank

hard

Fix the error in the code to push the next element of the same list into the min heap.

DSA Python

if element_idx + 1 < len(lists[list_idx]):
    heapq.heappush(min_heap, (lists[list_idx][[1]], list_idx, element_idx + 1))

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

Aelement_idx + 1

Belement_idx

Clist_idx

Delement_idx - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using the current element index instead of the next.

Using the list index instead of element index.

We push the next element at index 'element_idx + 1' from the same list into the heap.

4fill in blank

hard

Fill both blanks to append the smallest value to the result list and pop it from the heap.

DSA Python

val, list_idx, element_idx = heapq.[1](min_heap)
result.[2](val)

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

Aheappop

Bappend

Cpop

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'pop' instead of 'heappop' for the heap.

Using 'push' instead of 'append' for the list.

We pop the smallest element using 'heappop' and append its value to the result list.

5fill in blank

hard

Fill all three blanks to complete the merge function using a min heap.

DSA Python

import heapq

def merge_k_lists(lists):
    min_heap = []
    result = []
    for i in range(len(lists)):
        if lists[i]:
            heapq.heappush(min_heap, (lists[i][0], i, 0))
    while min_heap:
        val, list_idx, element_idx = heapq.[1](min_heap)
        result.[2](val)
        if element_idx + 1 < len(lists[list_idx]):
            heapq.heappush(min_heap, (lists[list_idx][[3]], list_idx, element_idx + 1))
    return result

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

Aheappop

Bappend

Celement_idx + 1

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'pop' instead of 'heappop' for the heap.

Appending the wrong variable to the result.

Using the wrong index for the next element.

We pop the smallest element with 'heappop', append its value, and push the next element at 'element_idx + 1'.

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