Merge K Sorted Lists Using Min Heap in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind using a min heap to merge K sorted lists?

A min heap helps efficiently find the smallest current element among all lists by always keeping the smallest element at the top, allowing us to build the merged list in sorted order.

beginner

How do you initialize the min heap when merging K sorted lists?

Insert the first element of each of the K sorted lists into the min heap to start the merging process.

intermediate

What happens after extracting the smallest element from the min heap during the merge?

After extracting the smallest element, insert the next element from the same list (where the extracted element came from) into the min heap, if it exists.

intermediate

Why is the time complexity of merging K sorted lists using a min heap O(N log K)?

Because each of the N total elements is pushed and popped from the min heap, and the heap size is at most K, each operation takes O(log K), resulting in O(N log K) overall.

beginner

What data structure is commonly used to represent the min heap in JavaScript for this problem?

A priority queue or a custom binary heap implemented as an array is commonly used to represent the min heap in JavaScript.

What is the first step when merging K sorted lists using a min heap?

A Insert the first element of each list into the min heap

B Merge all lists into one big list and sort it

C Extract the largest element from each list

D Reverse all lists

We start by inserting the first element of each list into the min heap to efficiently find the smallest element.

After extracting the smallest element from the min heap, what do you do next?

A Insert the largest element from another list

B Insert the next element from the same list into the min heap

C Remove all elements from the min heap

D Stop the process

We insert the next element from the same list to continue merging in sorted order.

What is the maximum size of the min heap during the merge of K sorted lists?

A N (total elements)

B N/K

C 1

D K

The min heap holds at most one element from each list, so its size is at most K.

What is the overall time complexity of merging K sorted lists using a min heap?

A O(N log K)

B O(N^2)

C O(K log N)

D O(N)

Each of the N elements is pushed and popped from a heap of size K, so the complexity is O(N log K).

Which data structure property is essential for the min heap in this problem?

A Elements are stored in reverse order

B The largest element is always at the top

C The smallest element is always at the top

D Elements are stored in insertion order

The min heap always keeps the smallest element at the top to efficiently find the next smallest element.

Explain step-by-step how to merge K sorted lists using a min heap.

Think about how the min heap helps pick the smallest element each time.

Describe why using a min heap is more efficient than merging all lists and sorting them at once.

Compare sorting all elements at once vs. incremental merging.

Practice

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

easy

A. To sort each individual list before merging

B. To efficiently find the smallest element among the heads of all lists

C. To store all elements from all lists at once

D. To reverse the order of elements in each list

Solution

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

   A min heap helps quickly find the smallest element among multiple candidates, which is essential when merging sorted lists.

2. Step 2: Apply this to merging K sorted lists

   By pushing the first element of each list into the min heap, we can always extract the smallest element efficiently to build the merged list.

3. Final Answer:

   To efficiently find the smallest element among the heads of all lists -> Option B

4. Quick Check:

   Min heap = smallest element finder [OK]

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

Common Mistakes:

• Thinking min heap sorts entire lists
• Assuming min heap stores all elements at once
• Confusing min heap with max heap

2. Which of the following is the correct way to insert an element into a min heap in JavaScript?

easy

A. heap.push(element); heapifyUp();

B. heap.pop(); heapifyDown();

C. heap.unshift(element); heapifyDown();

D. heap.shift(); heapifyUp();

Solution

1. Step 1: Understand insertion in min heap

   To insert, add the element at the end (push), then move it up to maintain heap property (heapifyUp).

2. Step 2: Match with options

   heap.push(element); heapifyUp(); correctly shows push followed by heapifyUp, which is the standard insertion method.

3. Final Answer:

   heap.push(element); heapifyUp(); -> Option A

4. Quick Check:

   Insert = push + heapifyUp [OK]

Hint: Insert by push then heapify up [OK]

Common Mistakes:

• Using pop or shift to insert
• Calling heapifyDown after insertion
• Using unshift which is inefficient

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

const lists = [[1,4,5],[1,3,4],[2,6]];
function mergeKLists(lists) {
  const minHeap = [];
  const result = [];
  for (let i = 0; i < lists.length; i++) {
    if (lists[i].length > 0) minHeap.push({ val: lists[i][0], listIndex: i, elementIndex: 0 });
  }
  minHeap.sort((a,b) => a.val - b.val);
  while (minHeap.length > 0) {
    const smallest = minHeap.shift();
    result.push(smallest.val);
    const nextIndex = smallest.elementIndex + 1;
    if (nextIndex < lists[smallest.listIndex].length) {
      minHeap.push({ val: lists[smallest.listIndex][nextIndex], listIndex: smallest.listIndex, elementIndex: nextIndex });
      minHeap.sort((a,b) => a.val - b.val);
    }
  }
  return result;
}
console.log(mergeKLists(lists));

medium

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

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

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

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

Solution

1. Step 1: Initialize min heap with first elements

   minHeap starts with {1,0,0}, {1,1,0}, {2,2,0} sorted as [1,1,2].

2. Step 2: Extract smallest and add next elements

   Extract 1 from list 0, add 4; extract 1 from list 1, add 3; extract 2 from list 2, add 6; continue until all elements merged.

3. Final Answer:

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

4. Quick Check:

   Merged sorted list = [1,1,2,3,4,4,5,6] [OK]

Hint: Track smallest and add next from same list [OK]

Common Mistakes:

• Not sorting minHeap after each insertion
• Forgetting to add next element from same list
• Returning concatenated lists instead of merged

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

function mergeKLists(lists) {
  const minHeap = [];
  const result = [];
  for (let i = 0; i < lists.length; i++) {
    minHeap.push({ val: lists[i][0], listIndex: i, elementIndex: 0 });
  }
  minHeap.sort((a,b) => a.val - b.val);
  while (minHeap.length > 0) {
    const smallest = minHeap.shift();
    result.push(smallest.val);
    const nextIndex = smallest.elementIndex + 1;
    if (nextIndex < lists[smallest.listIndex].length) {
      minHeap.push({ val: lists[smallest.listIndex][nextIndex], listIndex: smallest.listIndex, elementIndex: nextIndex });
      minHeap.sort((a,b) => a.val - b.val);
    }
  }
  return result;
}

medium

A. Not initializing minHeap with first elements of lists

B. Not pushing the first element of each list into minHeap

C. Using shift() instead of pop() to get smallest element

D. Missing sorting minHeap after pushing new element inside the loop

Solution

1. Step 1: Check minHeap maintenance inside loop

   After pushing a new element, minHeap must be sorted to keep smallest element at front.

2. Step 2: Identify missing sorting

   The code pushes new elements but does not sort minHeap again, breaking the min heap property.

3. Final Answer:

   Missing sorting minHeap after pushing new element inside the loop -> Option D

4. Quick Check:

   Sort after push to maintain min heap [OK]

Hint: Always re-sort min heap after insertion [OK]

Common Mistakes:

• Forgetting to sort minHeap after push
• Confusing shift() and pop() usage
• Not pushing initial elements into minHeap

5. Given K sorted linked lists, which approach best optimizes the time complexity when merging them using a min heap?

hard

A. Insert only the first element of each list into the min heap, then extract and insert next elements dynamically

B. Insert all elements from all lists into the min heap at once, then extract all

C. Merge lists two at a time without using a min heap

D. Concatenate all lists and sort the combined list

Solution

1. Step 1: Analyze time complexity of inserting all elements at once

   Inserting all elements at once leads to O(N log N) where N is total elements, which is less efficient.

2. Step 2: Understand dynamic insertion approach

   Inserting only first elements and dynamically adding next elements keeps heap size at K, leading to O(N log K) time, which is optimal.

3. Final Answer:

   Insert only the first element of each list into the min heap, then extract and insert next elements dynamically -> Option A

4. Quick Check:

   Dynamic insertion = O(N log K) best [OK]

Hint: Keep heap size small by dynamic insertion [OK]

Common Mistakes:

• Inserting all elements at once causing slow performance
• Ignoring heap size impact on complexity
• Using simple concatenation and sort which is inefficient