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

Time Complexity: Merge K Sorted Lists Using Min Heap

O(n log k)

Understanding Time Complexity

When merging multiple sorted lists, we want to know how the time needed grows as the number of lists and their sizes increase.

We ask: How does the merging time change when we have more lists or longer lists?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function mergeKLists(lists) {
  const minHeap = new MinHeap();
  for (const list of lists) {
    if (list) minHeap.insert(list);
  }
  const dummy = new ListNode(0);
  let current = dummy;
  while (!minHeap.isEmpty()) {
    const node = minHeap.extractMin();
    current.next = node;
    current = current.next;
    if (node.next) minHeap.insert(node.next);
  }
  return dummy.next;
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Extracting the smallest node from the min heap and inserting the next node from that list.
• How many times: This happens once for every node across all lists, so total nodes n times.

How Execution Grows With Input

Each node is inserted and extracted from the heap, which takes time depending on the number of lists k.

Input Size (n)	Approx. Operations
10 nodes, 3 lists	About 10 insertions and 10 extractions, each costing log 3 steps
100 nodes, 5 lists	About 100 insertions and 100 extractions, each costing log 5 steps
1000 nodes, 10 lists	About 1000 insertions and 1000 extractions, each costing log 10 steps

Pattern observation: The total work grows roughly with the total number of nodes times the log of the number of lists.

Final Time Complexity

Time Complexity: O(n log k)

This means the time grows with the total number of nodes n multiplied by the logarithm of the number of lists k.

Common Mistake

[X] Wrong: "Merging k lists is just like merging two lists, so it takes O(n) time."

[OK] Correct: Merging k lists involves repeatedly choosing the smallest among k current nodes, which costs extra time. The heap helps manage this efficiently, but the log k factor matters.

Interview Connect

Understanding this complexity shows you can handle multiple sorted inputs efficiently, a common real-world task when combining data streams or search results.

Self-Check

"What if we used a simple array instead of a min heap to find the smallest node 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 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