Merge K Sorted Lists Using Min Heap in DSA Rust - Execution Trace

Concept Flow - Merge K Sorted Lists Using Min Heap

Start with K sorted lists

↓

Create a min heap

↓

Insert first node of each list into min heap

↓

While min heap not empty

↓

Extract min node from heap

↓

Add extracted node to merged list

↓

If extracted node has next, insert next into heap

↓

Repeat until heap empty

↓

Return merged sorted list

We start by putting the first node of each sorted list into a min heap. Then we repeatedly take the smallest node from the heap, add it to the merged list, and if that node has a next node, we add that next node to the heap. This continues until all nodes are merged.

Execution Sample

DSA Rust

fn merge_k_lists(lists: Vec<Option<Box<ListNode>>>) -> Option<Box<ListNode>> {
    use std::cmp::Reverse;
    use std::collections::BinaryHeap;
    let mut heap: BinaryHeap<Reverse<(i32, usize)>> = BinaryHeap::new();
    // Insert first nodes (with list indices)
    // Extract min, add to result
    // Insert next nodes
    // Repeat
}

This code merges K sorted linked lists by using a min heap to always pick the smallest current node.

Execution Table

Step	Operation	Heap Content (values)	Extracted Node	Merged List State	Heap After Operation
1	Insert first nodes of each list	[1, 4, 5]	None	Empty	[1, 4, 5]
2	Extract min node	[1, 4, 5]	1	1 -> null	[4, 5]
3	Insert next of extracted node (1 -> 3)	[4, 5]	None	1 -> null	[3, 4, 5]
4	Extract min node	[3, 4, 5]	3	1 -> 3 -> null	[4, 5]
5	Insert next of extracted node (3 -> 4)	[4, 5]	None	1 -> 3 -> null	[4, 4, 5]
6	Extract min node	[4, 4, 5]	4	1 -> 3 -> 4 -> null	[4, 5]
7	Extract min node	[4, 5]	4	1 -> 3 -> 4 -> 4 -> null	[5]
8	Extract min node	[5]	5	1 -> 3 -> 4 -> 4 -> 5 -> null	[]
9	Heap empty, end	[]	None	1 -> 3 -> 4 -> 4 -> 5 -> null	[]

💡 Heap is empty, all nodes merged into one sorted list

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	Final
heap	[]	[1,4,5]	[4,5]	[3,4,5]	[4,5]	[4,4,5]	[4,5]	[5]	[]	[]
merged_list	Empty	Empty	1 -> null	1 -> null	1 -> 3 -> null	1 -> 3 -> null	1 -> 3 -> 4 -> null	1 -> 3 -> 4 -> 4 -> null	1 -> 3 -> 4 -> 4 -> 5 -> null	1 -> 3 -> 4 -> 4 -> 5 -> null

Key Moments - 3 Insights

Why do we insert only the first node of each list into the heap initially?

Because the lists are sorted, the first node is the smallest in each list. Inserting only these nodes ensures the heap always contains the next smallest candidates. This is shown in execution_table step 1.

Why do we insert the next node of the extracted node back into the heap?

After extracting the smallest node, its next node might be the next smallest in the merged list. Adding it to the heap keeps the heap updated with current smallest nodes. See steps 3 and 5 in execution_table.

When does the merging process stop?

It stops when the heap is empty, meaning all nodes from all lists have been extracted and merged. This is shown in step 9 of execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the heap content after step 3?

A [4, 5]

B [3, 4, 5]

C [1, 4, 5]

D [4, 4, 5]

💡 Hint

Check the 'Heap After Operation' column in row for step 3.

At which step does the merged list first contain the node with value 4?

A Step 6

B Step 5

C Step 4

D Step 7

💡 Hint

Look at the 'Merged List State' column and find when '4' appears first.

If the initial lists had no nodes, what would be the heap content after step 1?

A [0]

B [1, 4, 5]

C Empty

D Undefined

💡 Hint

Step 1 inserts first nodes of each list; no nodes means no insertions.

Concept Snapshot

Merge K Sorted Lists Using Min Heap:
- Insert first node of each list into min heap
- Extract min node from heap and add to merged list
- Insert next node of extracted node into heap if exists
- Repeat until heap empty
- Result is one merged sorted list

Full Transcript

To merge K sorted linked lists, we use a min heap to always pick the smallest current node. We start by inserting the first node of each list into the heap. Then, we repeatedly extract the smallest node, add it to the merged list, and if that node has a next node, we insert that next node into the heap. This process continues until the heap is empty, meaning all nodes are merged. The heap keeps track of the smallest candidates from each list, ensuring the merged list is sorted. Key points include inserting only the first nodes initially, adding next nodes after extraction, and stopping when the heap is empty.

Practice

1. What is the main purpose of using a min heap when merging K sorted lists?

easy

A. To reverse the order of elements in each list

B. To store all elements in sorted order before merging

C. To efficiently get the smallest element among the heads of all lists

D. To split the lists into smaller parts

Solution

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

   A min heap helps quickly find the smallest element among multiple sorted lists by always keeping the smallest element at the top.

2. Step 2: Why is this useful for merging K sorted lists?

   When merging, we repeatedly pick the smallest current element from all lists. The min heap allows this in O(log K) time, making the process efficient.

3. Final Answer:

   To efficiently get the smallest element among the heads of all lists -> Option C

4. Quick Check:

   Min heap = smallest element retrieval [OK]

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

Common Mistakes:

• Thinking min heap stores all elements sorted
• Confusing min heap with max heap
• Using min heap to split lists

2. Which Rust data structure is best suited to implement a min heap for merging K sorted lists?

easy

A. VecDeque

B. LinkedList

C. HashMap

D. BinaryHeap with Reverse wrapper

Solution

1. Step 1: Identify Rust's heap structure

   Rust's standard library provides BinaryHeap which is a max heap by default.

2. Step 2: How to get min heap behavior?

   By wrapping elements with Reverse, BinaryHeap behaves as a min heap, which is perfect for merging sorted lists.

3. Final Answer:

   BinaryHeap with Reverse wrapper -> Option D

4. Quick Check:

   BinaryHeap + Reverse = min heap [OK]

Hint: Use BinaryHeap with Reverse for min heap in Rust [OK]

Common Mistakes:

• Using VecDeque which is a queue, not a heap
• Choosing HashMap which is for key-value pairs
• Using LinkedList which is inefficient here

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

use std::collections::BinaryHeap;
use std::cmp::Reverse;

fn merge_k_lists(lists: Vec) -> Vec<i32> {
    let mut heap = BinaryHeap::new();
    for (i, list) in lists.iter().enumerate() {
        if !list.is_empty() {
            heap.push(Reverse((list[0], i, 0)));
        }
    }
    let mut result = Vec::new();
    while let Some(Reverse((val, list_i, idx))) = heap.pop() {
        result.push(val);
        if idx + 1 < lists[list_i].len() {
            heap.push(Reverse((lists[list_i][idx + 1], list_i, idx + 1)));
        }
    }
    result
}

fn main() {
    let lists = vec![vec![1,4,5], vec![1,3,4], vec![2,6]];
    let merged = merge_k_lists(lists);
    println!("{:?}", merged);
}

medium

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

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

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

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

Solution

1. Step 1: Initialize heap with first elements

   The heap starts with (1,0,0), (1,1,0), (2,2,0) from the three lists.

2. Step 2: Pop smallest and push next from same list

   Pop 1 from list 0, push 4; pop 1 from list 1, push 3; pop 2 from list 2, push 6; continue until all elements are merged.

3. Final Answer:

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

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:

• Ignoring duplicates
• Not pushing next element after pop
• Confusing max heap order with min heap

4. Identify the bug in this Rust code snippet for merging K sorted lists using a min heap:

use std::collections::BinaryHeap;
use std::cmp::Reverse;

fn merge_k_lists(lists: Vec) -> Vec<i32> {
    let mut heap = BinaryHeap::new();
    for (i, list) in lists.iter().enumerate() {
        heap.push(Reverse((list[0], i, 0)));
    }
    let mut result = Vec::new();
    while let Some(Reverse((val, list_i, idx))) = heap.pop() {
        result.push(val);
        if idx + 1 < lists[list_i].len() {
            heap.push(Reverse((lists[list_i][idx + 1], list_i, idx + 1)));
        }
    }
    result
}

medium

A. The code pushes first element without checking if list is empty

B. The heap should be a max heap, not min heap

C. The loop should use while instead of for

D. The result vector is not initialized

Solution

1. Step 1: Check for empty lists before pushing

   The code pushes list[0] without checking if the list is empty, which can cause a panic.

2. Step 2: Fix by adding a check

   Adding if !list.is_empty() before pushing prevents runtime errors.

3. Final Answer:

   The code pushes first element without checking if list is empty -> Option A

4. Quick Check:

   Empty list check needed before accessing list[0] [OK]

Hint: Always check list is not empty before accessing index 0 [OK]

Common Mistakes:

• Ignoring empty lists causing panic
• Confusing min heap with max heap
• For loop usage is correct here

5. Given K sorted linked lists, how would you modify the min heap merge approach to handle lists with millions of elements efficiently in Rust?

hard

A. Load all elements into memory before merging

B. Use iterators for lazy traversal and push only current nodes into the min heap

C. Use a max heap instead of a min heap for better performance

D. Merge lists pairwise without using any heap

Solution

1. Step 1: Understand memory constraints with large lists

   Loading millions of elements at once can cause memory overflow or slow performance.

2. Step 2: Use iterators for lazy traversal

   Using iterators allows processing one node at a time, pushing only current nodes into the min heap, saving memory and improving efficiency.

3. Step 3: Why other options are less efficient

   Loading all elements wastes memory, max heap is incorrect for smallest element retrieval, and pairwise merging is slower for large K.

4. Final Answer:

   Use iterators for lazy traversal and push only current nodes into the min heap -> Option B

5. Quick Check:

   Lazy iterator + min heap = efficient large list merge [OK]

Hint: Use iterators to avoid loading all elements at once [OK]

Common Mistakes:

• Loading entire lists into memory
• Using max heap instead of min heap
• Ignoring performance impact of pairwise merges