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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the min heap as an empty array.

DSA Javascript

const minHeap = [1];

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

A0

B[]

Cnull

D{}

Attempts:

3 left

💡 Hint

Common Mistakes

Using an object {} instead of an array.

Initializing with null or 0.

We use an empty array [] to represent the min heap initially.

2fill in blank

medium

Complete the code to push the first node of each list into the min heap.

DSA Javascript

for (let i = 0; i < lists.length; i++) {
  if (lists[i]) {
    minHeap.push({ val: lists[i].val, listIndex: i, node: lists[i] });
  }
}

// Now build the heap
for (let i = Math.floor(minHeap.length / 2) - 1; i >= 0; i--) {
  [1](minHeap, i);
}

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

Apop

Bsort

Cpush

Dheapify

Attempts:

3 left

💡 Hint

Common Mistakes

Using sort instead of heapify.

Calling push or pop incorrectly.

We call heapify to build the min heap from the array.

3fill in blank

hard

Fix the error in the function that extracts the minimum element from the heap.

DSA Javascript

function extractMin(heap) {
  if (heap.length === 0) return null;
  const min = heap[0];
  heap[0] = heap[heap.length - 1];
  heap.pop();
  [1](heap, 0);
  return min;
}

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

Aheapify

Bpush

Cextract

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Calling pop or push instead of heapify.

Using extract which is not defined.

After removing the root, we call heapify to restore the min heap property.

4fill in blank

hard

Fill both blanks to insert the next node from the extracted list into the heap and maintain the heap property.

DSA Javascript

if (min.node.next) {
  minHeap.push({ val: min.node.next.val, listIndex: min.listIndex, node: min.node.next });
  [1](minHeap, minHeap.length - 1);
}

function [2](heap, index) {
  // heapifyDown implementation
}

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

AheapifyUp

BheapifyDown

Cheapify

DheapifyUpwards

Attempts:

3 left

💡 Hint

Common Mistakes

Using heapify instead of heapifyUp after push.

Confusing heapifyUp and heapifyDown.

We call heapifyUp after pushing a new element to fix upwards, and heapifyDown is the function that fixes downwards.

5fill in blank

hard

Fill all three blanks to complete the main loop that extracts the smallest node and appends it to the merged list.

DSA Javascript

while (minHeap.length > 0) {
  const min = [1](minHeap);
  current.next = min.node;
  current = current.next;

  if (min.node.next) {
    minHeap.push({ val: min.node.next.val, listIndex: min.listIndex, node: min.node.next });
    [2](minHeap, minHeap.length - 1);
  }

  [3](minHeap, 0);
}

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

AextractMin

BheapifyUp

CheapifyDown

DpopMin

Attempts:

3 left

💡 Hint

Common Mistakes

Using popMin which is undefined.

Mixing heapifyUp and heapifyDown order.

We extract the minimum node with extractMin, fix the heap upwards after pushing a new node with heapifyUp, and fix downwards with heapifyDown.

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