Merge K Sorted Lists Using Min Heap in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of merging 3 sorted lists using min heap

What is the printed state of the merged linked list after merging these 3 sorted lists using a min heap?

DSA Rust

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

#[derive(Debug, PartialEq, Eq, PartialOrd, Ord)]
struct ListNode {
    val: i32,
    next: Option<Box<ListNode>>,
}

impl ListNode {
    fn new(val: i32) -> Self {
        ListNode { val, next: None }
    }
}

fn merge_k_lists(lists: Vec<Option<Box<ListNode>>>) -> Option<Box<ListNode>> {
    let mut heap = BinaryHeap::new();

    for (i, list) in lists.into_iter().enumerate() {
        if let Some(node) = list {
            heap.push(Reverse((node.val, i, node)));
        }
    }

    let mut dummy = Box::new(ListNode::new(0));
    let mut tail = &mut dummy;

    while let Some(Reverse((val, i, mut node))) = heap.pop() {
        tail.next = Some(Box::new(ListNode::new(val)));
        tail = tail.next.as_mut().unwrap();

        if let Some(next_node) = node.next.take() {
            heap.push(Reverse((next_node.val, i, next_node)));
        }
    }

    dummy.next
}

fn main() {
    let list1 = Some(Box::new(ListNode { val: 1, next: Some(Box::new(ListNode { val: 4, next: Some(Box::new(ListNode::new(5))) })) }));
    let list2 = Some(Box::new(ListNode { val: 1, next: Some(Box::new(ListNode { val: 3, next: Some(Box::new(ListNode::new(4))) })) }));
    let list3 = Some(Box::new(ListNode { val: 2, next: Some(Box::new(ListNode::new(6))) }));

    let merged = merge_k_lists(vec![list1, list2, list3]);

    let mut current = &merged;
    while let Some(node) = current {
        print!("{} -> ", node.val);
        current = &node.next;
    }
    print!("null\n");
}

A 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null

B 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> null

C 1 -> 1 -> 3 -> 4 -> 4 -> 5 -> 6 -> null

D 1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> null

💡 Hint

Remember that the min heap always picks the smallest current node from all lists.

The merged list contains all nodes from the three sorted lists in ascending order. The min heap helps pick the smallest node at each step, resulting in the sequence 1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> null.

🧠 Conceptual

intermediate

Why use a min heap to merge K sorted lists?

What is the main advantage of using a min heap to merge K sorted linked lists compared to merging them one by one?

A It reduces the overall time complexity by always selecting the smallest current node efficiently.

B It uses less memory by storing only one list at a time.

C It avoids using any extra data structures.

D It sorts each individual list before merging.

💡 Hint

Think about how the min heap helps pick the smallest element quickly.

Using a min heap allows efficient selection of the smallest node among all current heads of the K lists, reducing the time complexity from O(NK) to O(N log K), where N is total nodes.

🔧 Debug

advanced

Identify the error in this Rust code merging K sorted lists

What error will this Rust code produce when trying to compile?

DSA Rust

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

struct ListNode {
    val: i32,
    next: Option<Box<ListNode>>,
}

fn merge_k_lists(lists: Vec<Option<Box<ListNode>>>) -> Option<Box<ListNode>> {
    let mut heap = BinaryHeap::new();

    for list in lists {
        if let Some(node) = list {
            heap.push(Reverse((node.val, node)));
        }
    }

    let mut dummy = Box::new(ListNode { val: 0, next: None });
    let mut tail = &mut dummy;

    while let Some(Reverse((val, mut node))) = heap.pop() {
        tail.next = Some(Box::new(ListNode { val, next: None }));
        tail = tail.next.as_mut().unwrap();

        if let Some(next_node) = node.next.take() {
            heap.push(Reverse((next_node.val, next_node)));
        }
    }

    dummy.next
}

A error[E0382]: use of moved value `node`

B error[E0277]: the trait `Ord` is not implemented for `(i32, Box<ListNode>)`

C error[E0425]: cannot find value `heap` in this scope

D error[E0599]: no method named `take` found for `Option<Box<ListNode>>`

💡 Hint

Check if the tuple pushed into the heap implements the required traits.

BinaryHeap requires its elements to implement Ord. The tuple (i32, Box) does not implement Ord because Box does not implement Ord, causing a compilation error.

📝 Syntax

advanced

Which option correctly uses a min heap to merge K sorted lists in Rust?

Select the option that correctly pushes nodes into a min heap for merging K sorted lists.

DSA Rust

struct ListNode {
    val: i32,
    next: Option<Box<ListNode>>,
}

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

fn merge_k_lists(lists: Vec<Option<Box<ListNode>>>) {
    let mut heap = BinaryHeap::new();
    // Push nodes here
}

A

for list in lists {
    if let Some(node) = list {
        heap.push(Reverse((node.val, node)));
    }
}

B

for list in lists {
    if let Some(node) = list {
        heap.push(node.val);
    }
}

C

for (i, list) in lists.into_iter().enumerate() {
    if let Some(node) = list {
        heap.push(Reverse((node.val, i, node)));
    }
}

D

for list in lists {
    heap.push((list.val, list));
}

💡 Hint

Remember to include an index to break ties and use Reverse for min heap.

Option C correctly pushes a tuple with value, index, and node wrapped in Reverse to create a min heap. Option C misses the index causing Ord issues. Options C and D are invalid syntax or incomplete.

🚀 Application

expert

What is the time complexity of merging K sorted lists with total N nodes using a min heap?

Given K sorted linked lists with a total of N nodes, what is the time complexity of merging them using a min heap?

A O(N log K)

B O(K log N)

C O(N K)

D O(N + K)

💡 Hint

Consider how many times nodes are pushed and popped from the heap and the heap size.

Each of the N nodes is pushed and popped from a heap of size at most K, so each operation costs O(log K), resulting in O(N log K) time complexity.

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