Merge K Sorted Lists Using Min Heap in DSA Rust - Build from Scratch

Merge K Sorted Lists Using Min Heap

📖 Scenario: Imagine you have several sorted lists of numbers from different friends. You want to combine all these lists into one big sorted list quickly and easily.

🎯 Goal: You will build a program in Rust that merges multiple sorted lists into one sorted list using a min heap. This helps you get the combined sorted list efficiently.

📋 What You'll Learn

Create a vector of vectors called sorted_lists with exact sorted integer lists

Create a min heap using BinaryHeap with reversed order to get smallest elements first

Implement the merging logic using the min heap to combine all lists into one sorted vector

Print the final merged sorted list

💡 Why This Matters

🌍 Real World

Merging sorted lists is useful in combining data from multiple sources like logs, search results, or streams in sorted order.

💼 Career

Understanding min heaps and merging techniques is important for software engineers working on data processing, databases, and performance optimization.

1

Create the sorted lists

Create a variable called sorted_lists as a vector of vectors with these exact sorted integer lists: [1, 4, 5], [1, 3, 4], and [2, 6].

DSA Rust

# Create the sorted_lists variable with the given lists
# Your code here

Hint

Use vec![...] to create vectors inside a vector.

2

Set up the min heap and helper variables

Add these lines below the sorted_lists declaration: create a mutable BinaryHeap called min_heap that stores tuples of (i32, usize, usize) with reversed order using std::cmp::Reverse. Also create a mutable vector called result to store the merged numbers.

DSA Rust

let sorted_lists = vec![vec![1, 4, 5], vec![1, 3, 4], vec![2, 6]];
// Create min_heap and result vector here
# Your code here

Hint

Import BinaryHeap and Reverse from std::collections and std::cmp.

3

Push first elements of each list into the min heap

Use a for loop with variable i over the range 0..sorted_lists.len(). Inside the loop, if the list at index i is not empty, push a Reverse tuple containing the first element of that list, the list index i, and element index 0 into min_heap.

DSA Rust

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

let sorted_lists = vec![vec![1, 4, 5], vec![1, 3, 4], vec![2, 6]];

let mut min_heap: BinaryHeap> = BinaryHeap::new();
let mut result: Vec = Vec::new();

// Push first elements of each list into min_heap
# Your code here

Hint

Check if each list is not empty before pushing its first element.

4

Merge all lists and print the result

Use a while loop that runs while min_heap is not empty. Inside the loop, pop the smallest element from min_heap into a variable Reverse((val, list_idx, elem_idx)). Push val into result. If the next element index elem_idx + 1 is less than the length of the list at list_idx, push the next element from that list into min_heap as a Reverse tuple. Finally, print the result vector.

DSA Rust

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

let sorted_lists = vec![vec![1, 4, 5], vec![1, 3, 4], vec![2, 6]];

let mut min_heap: BinaryHeap> = BinaryHeap::new();
let mut result: Vec = Vec::new();

for i in 0..sorted_lists.len() {
    if !sorted_lists[i].is_empty() {
        min_heap.push(Reverse((sorted_lists[i][0], i, 0)));
    }
}

// Merge lists using min_heap and print result
# Your code here

Hint

Use while let Some(Reverse((val, list_idx, elem_idx))) = min_heap.pop() to pop elements safely.

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