Merge K Sorted Lists Using Min Heap in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a min heap using Rust's BinaryHeap with reversed order.

DSA Rust

let mut heap = std::collections::BinaryHeap::new();
heap.[1](5);

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Apop

Bpush

Cpeek

Dnew

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push to add elements.

Trying to create a new heap inside the push call.

The push method adds an element to the heap. Here, we add the value 5 to the min heap.

2fill in blank

medium

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

DSA Rust

if let Some(smallest) = heap.[1]() {
    println!("Smallest: {}", smallest);
}

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Apop

Bpeek

Cremove

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek instead of pop, which does not remove the element.

Using push instead of pop.

The pop method removes and returns the smallest element from the min heap.

3fill in blank

hard

Fix the error in the code to correctly compare elements in the min heap by reversing the order.

DSA Rust

use std::cmp::[1];

struct Wrapper(i32);

impl PartialEq for Wrapper {
    fn eq(&self, other: &Self) -> bool {
        self.0 == other.0
    }
}

impl Eq for Wrapper {}

impl PartialOrd for Wrapper {
    fn partial_cmp(&self, other: &Self) -> Option<std::cmp::Ordering> {
        Some(self.cmp(other))
    }
}

impl Ord for Wrapper {
    fn cmp(&self, other: &Self) -> std::cmp::Ordering {
        other.0.[2](self.0)
    }
}

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AOrdering

BReverse

CPartialOrd

Dcmp

Attempts:

3 left

💡 Hint

Common Mistakes

Importing the wrong trait or enum.

Not reversing the order in cmp method.

The code uses std::cmp::Ordering for comparison and reverses the order by calling other.0.cmp(self.0) to create a min heap behavior.

4fill in blank

hard

Fill both blanks to correctly pop the smallest node and push the next node from the list.

DSA Rust

while let Some(Wrapper(node)) = heap.[1]() {
    if node.next[2] {
        heap.push(Wrapper(node.next.unwrap()));
    }
    // process node.val
}

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Apop

Bpeek

Cis_some()

D== None

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek instead of pop to remove the node.

Checking next == None instead of is_some().

We pop the smallest node from the heap and check if the next node exists using is_some() before pushing it.

5fill in blank

hard

Fill both blanks to build the merged list by popping from the heap and linking nodes.

DSA Rust

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

while let Some(Wrapper(node)) = heap.[1]() {
    tail.next = Some(node);
    tail = tail.next.as_mut().unwrap();
    if tail.next[2] {
        heap.push(Wrapper(tail.next.unwrap()));
    }
}

return dummy.next;

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

Apop

Bis_some()

C== None

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek instead of pop to get nodes.

Checking next == None instead of is_some().

We pop the smallest node from the heap, link it to the merged list, then if the next node exists (is_some()), we push it to the heap.

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