K-way merge with heaps in Data Structures Theory - Step-by-Step Execution

Concept Flow - K-way merge with heaps

Start with K sorted lists

↓

Insert first element of each list into min-heap

↓

Extract min element from heap

↓

Add extracted element to merged output

↓

Insert next element from same list into heap if exists

↓

Heap empty?

No→Repeat extract and insert

Yes↓

Done: merged sorted list

Start by putting the first element of each sorted list into a min-heap. Then repeatedly extract the smallest element, add it to the output, and insert the next element from the same list until all lists are merged.

Execution Sample

Data Structures Theory

lists = [[1,4,7],[2,5,8],[3,6,9]]
heap = []
for i in range(len(lists)):
    heapq.heappush(heap, (lists[i][0], i, 0))
while heap:
    val, list_i, elem_i = heapq.heappop(heap)

This code merges 3 sorted lists by pushing their first elements into a heap, then repeatedly popping the smallest and pushing the next element from the same list.

Analysis Table

Step	Operation	Heap Content (value, list_index, element_index)	Extracted Element	Merged Output	Next Inserted Element
1	Insert first elements	[(1,0,0),(2,1,0),(3,2,0)]
2	Extract min	[(2,1,0),(3,2,0)]	(1,0,0)	[1]	(4,0,1) inserted
3	Insert next element from list 0	[(2,1,0),(3,2,0),(4,0,1)]		[1]
4	Extract min	[(3,2,0),(4,0,1)]	(2,1,0)	[1,2]	(5,1,1) inserted
5	Insert next element from list 1	[(3,2,0),(4,0,1),(5,1,1)]		[1,2]
6	Extract min	[(4,0,1),(5,1,1)]	(3,2,0)	[1,2,3]	(6,2,1) inserted
7	Insert next element from list 2	[(4,0,1),(5,1,1),(6,2,1)]		[1,2,3]
8	Extract min	[(5,1,1),(6,2,1)]	(4,0,1)	[1,2,3,4]	(7,0,2) inserted
9	Insert next element from list 0	[(5,1,1),(6,2,1),(7,0,2)]		[1,2,3,4]
10	Extract min	[(6,2,1),(7,0,2)]	(5,1,1)	[1,2,3,4,5]	(8,1,2) inserted
11	Insert next element from list 1	[(6,2,1),(7,0,2),(8,1,2)]		[1,2,3,4,5]
12	Extract min	[(7,0,2),(8,1,2)]	(6,2,1)	[1,2,3,4,5,6]	(9,2,2) inserted
13	Insert next element from list 2	[(7,0,2),(8,1,2),(9,2,2)]		[1,2,3,4,5,6]
14	Extract min	[(8,1,2),(9,2,2)]	(7,0,2)	[1,2,3,4,5,6,7]	No next element
15	Extract min	[(9,2,2)]	(8,1,2)	[1,2,3,4,5,6,7,8]	No next element
16	Extract min	[]	(9,2,2)	[1,2,3,4,5,6,7,8,9]	No next element

💡 Heap is empty after extracting all elements, merge complete.

State Tracker

Variable	Start	After Step 1	After Step 4	After Step 8	After Step 12	Final
heap	[]	[(1,0,0),(2,1,0),(3,2,0)]	[(3,2,0),(4,0,1)]	[(4,0,1),(5,1,1),(6,2,1)]	[(7,0,2),(8,1,2),(9,2,2)]	[]
merged_output	[]	[]	[1,2]	[1,2,3,4]	[1,2,3,4,5,6]	[1,2,3,4,5,6,7,8,9]

Key Insights - 3 Insights

Why do we insert the next element from the same list after extracting an element?

Because each extracted element comes from a specific list, inserting the next element from that list keeps the heap updated with the smallest remaining elements, ensuring the merged output stays sorted (see execution_table steps 2, 4, 6, etc.).

What happens when a list runs out of elements?

When there is no next element in the list, we do not insert anything into the heap for that list. This reduces the heap size gradually until all elements are merged (see execution_table steps 14-16).

Why use a min-heap for merging?

A min-heap efficiently gives the smallest element among all current candidates from the lists, allowing us to merge in sorted order without scanning all lists repeatedly (see concept_flow and heap content in execution_table).

Visual Quiz - 3 Questions

Look at the execution_table at step 6. What is the heap content after inserting the next element from list 2?

A [(3,2,0),(4,0,1),(5,1,1)]

B [(4,0,1),(5,1,1),(6,2,1)]

C [(2,1,0),(3,2,0),(4,0,1)]

D [(5,1,1),(6,2,1),(7,0,2)]

💡 Hint

Check the 'Heap Content' column at step 6 in the execution_table.

At which step does the heap become empty, ending the merge?

A Step 14

B Step 15

C Step 16

D Step 13

💡 Hint

Look at the 'Heap Content' column and find when it becomes [].

If the first element of list 1 was 0 instead of 2, how would the first extracted element change?

A It would still be (1,0,0)

B It would be (0,1,0)

C It would be (2,1,0)

D It would be (3,2,0)

💡 Hint

The heap always extracts the smallest value; check the initial heap content in execution_table step 1.

Concept Snapshot

K-way merge with heaps:
- Start by pushing first element of each sorted list into a min-heap.
- Extract the smallest element from the heap and add to output.
- Insert the next element from the same list into the heap if available.
- Repeat until heap is empty.
- Efficiently merges K sorted lists into one sorted list in O(N log K) time.

Full Transcript

K-way merge with heaps merges multiple sorted lists by using a min-heap to always pick the smallest current element. Initially, the first element of each list is inserted into the heap. Then, the smallest element is extracted and added to the merged output. The next element from the same list is inserted into the heap if it exists. This process repeats until the heap is empty, meaning all elements have been merged. This method efficiently merges K sorted lists by maintaining a heap of size at most K, ensuring the next smallest element is always accessible.

Practice

1. What is the main purpose of using a min-heap in a K-way merge algorithm?

easy

A. To store all elements in a single large array

B. To sort each individual list before merging

C. To reverse the order of elements in each list

D. To efficiently find and extract the smallest element among multiple sorted lists

Solution

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

   A min-heap helps quickly find the smallest element among all current candidates from each list.

2. Step 2: Connect min-heap usage to K-way merge

   By always extracting the smallest element, the algorithm efficiently merges sorted lists without scanning all elements repeatedly.

3. Final Answer:

   To efficiently find and extract the smallest element among multiple sorted lists -> Option D

4. Quick Check:

   Min-heap = smallest element extraction [OK]

Hint: Min-heap always gives smallest element fast in merging [OK]

Common Mistakes:

• Thinking min-heap sorts individual lists
• Assuming min-heap stores all elements at once
• Confusing min-heap with max-heap

2. Which of the following is the correct initial step when implementing a K-way merge using a min-heap?

easy

A. Remove the largest element from each list

B. Insert all elements from all lists into the min-heap at once

C. Insert the first element of each sorted list into the min-heap

D. Sort each list again before merging

Solution

1. Step 1: Identify the initial heap population

   The algorithm starts by inserting the first element of each sorted list into the min-heap to track the smallest candidates.

2. Step 2: Explain why not all elements are inserted initially

   Inserting all elements at once would be inefficient and defeat the purpose of incremental merging.

3. Final Answer:

   Insert the first element of each sorted list into the min-heap -> Option C

4. Quick Check:

   Start heap with first elements only [OK]

Hint: Heap starts with first elements, not all at once [OK]

Common Mistakes:

• Inserting all elements at once causing inefficiency
• Sorting lists again unnecessarily
• Removing largest elements instead of smallest

3. Consider three sorted lists: [1, 4, 7], [2, 5, 8], and [3, 6, 9]. Using a K-way merge with a min-heap, what is the first element extracted from the heap?

medium

A. 1

B. 2

C. 3

D. 4

Solution

1. Step 1: Insert first elements of each list into the min-heap

   The heap initially contains 1, 2, and 3 from the three lists.

2. Step 2: Extract the smallest element from the heap

   The smallest among 1, 2, and 3 is 1, so it is extracted first.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Smallest first element = 1 [OK]

Hint: First extracted is smallest first element from all lists [OK]

Common Mistakes:

• Choosing second or third smallest element first
• Confusing heap order with list order
• Ignoring initial heap contents

4. You implemented a K-way merge with heaps but the merged output is not sorted. What is the most likely mistake?

medium

A. Using a max-heap instead of a min-heap

B. Not inserting the next element from a list after extracting its smallest element

C. Sorting each list before merging

D. Inserting duplicate elements into the heap

Solution

1. Step 1: Understand the heap update process in K-way merge

   After extracting the smallest element from a list, the next element from that list must be inserted into the heap to maintain correct order.

2. Step 2: Identify the effect of missing this step

   If the next element is not inserted, the heap misses candidates, causing the merged output to be unsorted or incomplete.

3. Final Answer:

   Not inserting the next element from a list after extracting its smallest element -> Option B

4. Quick Check:

   Insert next element after extraction [OK]

Hint: Always add next element after extraction to keep order [OK]

Common Mistakes:

• Using max-heap which reverses order
• Sorting lists again unnecessarily
• Thinking duplicates cause sorting errors

5. You have 4 sorted lists of different lengths: [1, 3, 5], [2, 4], [0, 6, 7, 8], and [9]. Using a K-way merge with a min-heap, what is the correct order of elements extracted?

hard

A. [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

B. [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]

C. [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

D. [0, 2, 1, 3, 4, 5, 6, 7, 8, 9]

Solution

1. Step 1: Insert first elements of all lists into the min-heap

   Heap starts with 1, 2, 0, and 9.

2. Step 2: Extract elements in ascending order, inserting next from each list

   Extract 0, insert 6; extract 1, insert 3; extract 2, insert 4; extract 3, insert 5; extract 4, no next; extract 5, no next; extract 6, insert 7; extract 7, insert 8; extract 8, no next; extract 9, no next.

3. Final Answer:

   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] -> Option A

4. Quick Check:

   Sorted merge order = ascending combined list [OK]

Hint: Extract smallest, insert next from same list until all done [OK]

Common Mistakes:

• Ignoring list lengths causing wrong order
• Extracting elements out of heap order
• Confusing max-heap with min-heap results