Merge K Sorted Lists Using Min Heap in DSA Rust - Time & Space Complexity

Time Complexity: Merge K Sorted Lists Using Min Heap

O(n log k)

Understanding Time Complexity

We want to know how the time needed to merge multiple sorted lists grows as we add more lists or elements.

How does using a min heap help us merge efficiently?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

fn merge_k_lists(lists: Vec>) -> Vec {
    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, elem_i))) = heap.pop() {
        result.push(val);
        if elem_i + 1 < lists[list_i].len() {
            heap.push(Reverse((lists[list_i][elem_i + 1], list_i, elem_i + 1)));
        }
    }
    result
}
    

This code merges k sorted lists by always picking the smallest current element using a min heap.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Extracting the smallest element from the heap and pushing the next element from the same list.
• How many times: Once for every element across all lists combined (total n elements).

How Execution Grows With Input

Each element is pushed and popped from the heap once. The heap size is at most k (number of lists).

Input Size (n)	Approx. Operations
10 (e.g., 2 lists)	About 10 push/pop operations on heap of size ≤ 2
100 (e.g., 5 lists)	About 100 push/pop operations on heap of size ≤ 5
1000 (e.g., 10 lists)	About 1000 push/pop operations on heap of size ≤ 10

Pattern observation: Operations grow linearly with total elements, but each heap operation depends on k, which is usually much smaller than n.

Final Time Complexity

Time Complexity: O(n log k)

This means the time grows mostly with the total number of elements, but each step is faster because the heap size depends on the number of lists, not elements.

Common Mistake

[X] Wrong: "Merging k lists always takes O(n * k) time because we compare all lists each time."

[OK] Correct: Using a min heap keeps track of the smallest elements efficiently, so we only do about log k work per element, not k.

Interview Connect

Understanding this time complexity shows you can use the right data structure to handle multiple sorted inputs efficiently, a skill useful in many coding challenges and real-world tasks.

Self-Check

"What if we used a simple array instead of a min heap to find the smallest element each time? How would the time complexity change?"

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